auto commit
This commit is contained in:
CyC2018
@ -1,146 +1,134 @@
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<!-- GFM-TOC -->
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* [快速选择](#快速选择)
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* [堆排序](#堆排序)
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* [Kth Element](#kth-element)
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* [桶排序](#桶排序)
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* [出现频率最多的 k 个数](#出现频率最多的-k-个数)
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* [按照字符出现次数对字符串排序](#按照字符出现次数对字符串排序)
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* [荷兰国旗问题](#荷兰国旗问题)
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* [按颜色进行排序](#按颜色进行排序)
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<!-- GFM-TOC -->
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# 快速选择
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用于求解 **Kth Element** 问题，使用快速排序的 partition() 进行实现。
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# 快速选择
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需要先打乱数组，否则最坏情况下时间复杂度为 O(N<sup>2</sup>)。
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用于求解 **Kth Element** 问题，使用快速排序的 partition() 进行实现。
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# 堆排序
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需要先打乱数组，否则最坏情况下时间复杂度为 O(N<sup>2</sup>)。
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用于求解 **TopK Elements** 问题，通过维护一个大小为 K 的堆，堆中的元素就是 TopK Elements。
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# 堆排序
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堆排序也可以用于求解 Kth Element 问题，堆顶元素就是 Kth Element。
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用于求解 **TopK Elements** 问题，通过维护一个大小为 K 的堆，堆中的元素就是 TopK Elements。
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快速选择也可以求解 TopK Elements 问题，因为找到 Kth Element 之后，再遍历一次数组，所有小于等于 Kth Element 的元素都是 TopK Elements。
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堆排序也可以用于求解 Kth Element 问题，堆顶元素就是 Kth Element。
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可以看到，快速选择和堆排序都可以求解 Kth Element 和 TopK Elements 问题。
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快速选择也可以求解 TopK Elements 问题，因为找到 Kth Element 之后，再遍历一次数组，所有小于等于 Kth Element 的元素都是 TopK Elements。
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## Kth Element
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可以看到，快速选择和堆排序都可以求解 Kth Element 和 TopK Elements 问题。
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[215. Kth Largest Element in an Array (Medium)](https://leetcode.com/problems/kth-largest-element-in-an-array/description/)
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## Kth Element
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题目描述：找到第 k 大的元素。
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[215. Kth Largest Element in an Array (Medium)](https://leetcode.com/problems/kth-largest-element-in-an-array/description/)
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题目描述：找到第 k 大的元素。
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**排序** ：时间复杂度 O(NlogN)，空间复杂度 O(1)
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**排序**：时间复杂度 O(NlogN)，空间复杂度 O(1)
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```java
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public int findKthLargest(int[] nums, int k) {
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Arrays.sort(nums);
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return nums[nums.length - k];
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public int findKthLargest(int[] nums, int k) {
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Arrays.sort(nums);
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return nums[nums.length - k];
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}
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```
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**堆排序** ：时间复杂度 O(NlogK)，空间复杂度 O(K)。
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**堆排序**：时间复杂度 O(NlogK)，空间复杂度 O(K)。
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```java
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public int findKthLargest(int[] nums, int k) {
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PriorityQueue<Integer> pq = new PriorityQueue<>(); // 小顶堆
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for (int val : nums) {
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pq.add(val);
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if (pq.size() > k) // 维护堆的大小为 K
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pq.poll();
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}
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return pq.peek();
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public int findKthLargest(int[] nums, int k) {
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PriorityQueue<Integer> pq = new PriorityQueue<>(); // 小顶堆
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for (int val : nums) {
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pq.add(val);
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if (pq.size() > k) // 维护堆的大小为 K
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pq.poll();
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}
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return pq.peek();
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}
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```
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**快速选择** ：时间复杂度 O(N)，空间复杂度 O(1)
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**快速选择**：时间复杂度 O(N)，空间复杂度 O(1)
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```java
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public int findKthLargest(int[] nums, int k) {
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k = nums.length - k;
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int l = 0, h = nums.length - 1;
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while (l < h) {
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int j = partition(nums, l, h);
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if (j == k) {
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break;
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} else if (j < k) {
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l = j + 1;
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} else {
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h = j - 1;
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}
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}
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return nums[k];
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public int findKthLargest(int[] nums, int k) {
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k = nums.length - k;
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int l = 0, h = nums.length - 1;
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while (l < h) {
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int j = partition(nums, l, h);
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if (j == k) {
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break;
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} else if (j < k) {
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l = j + 1;
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} else {
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h = j - 1;
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}
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}
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return nums[k];
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}
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private int partition(int[] a, int l, int h) {
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int i = l, j = h + 1;
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while (true) {
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while (a[++i] < a[l] && i < h) ;
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while (a[--j] > a[l] && j > l) ;
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if (i >= j) {
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break;
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}
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swap(a, i, j);
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}
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swap(a, l, j);
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return j;
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private int partition(int[] a, int l, int h) {
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int i = l, j = h + 1;
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while (true) {
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while (a[++i] < a[l] && i < h) ;
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while (a[--j] > a[l] && j > l) ;
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if (i >= j) {
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break;
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}
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swap(a, i, j);
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}
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swap(a, l, j);
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return j;
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}
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private void swap(int[] a, int i, int j) {
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int t = a[i];
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a[i] = a[j];
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a[j] = t;
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private void swap(int[] a, int i, int j) {
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int t = a[i];
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a[i] = a[j];
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a[j] = t;
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}
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```
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# 桶排序
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# 桶排序
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## 出现频率最多的 k 个数
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## 出现频率最多的 k 个数
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[347. Top K Frequent Elements (Medium)](https://leetcode.com/problems/top-k-frequent-elements/description/)
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[347. Top K Frequent Elements (Medium)](https://leetcode.com/problems/top-k-frequent-elements/description/)
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```html
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Given [1,1,1,2,2,3] and k = 2, return [1,2].
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Given [1,1,1,2,2,3] and k = 2, return [1,2].
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```
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设置若干个桶，每个桶存储出现频率相同的数，并且桶的下标代表桶中数出现的频率，即第 i 个桶中存储的数出现的频率为 i。
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设置若干个桶，每个桶存储出现频率相同的数，并且桶的下标代表桶中数出现的频率，即第 i 个桶中存储的数出现的频率为 i。
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把数都放到桶之后，从后向前遍历桶，最先得到的 k 个数就是出现频率最多的的 k 个数。
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把数都放到桶之后，从后向前遍历桶，最先得到的 k 个数就是出现频率最多的的 k 个数。
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```java
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public List<Integer> topKFrequent(int[] nums, int k) {
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Map<Integer, Integer> frequencyForNum = new HashMap<>();
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for (int num : nums) {
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frequencyForNum.put(num, frequencyForNum.getOrDefault(num, 0) + 1);
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}
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List<Integer>[] buckets = new ArrayList[nums.length + 1];
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for (int key : frequencyForNum.keySet()) {
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int frequency = frequencyForNum.get(key);
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if (buckets[frequency] == null) {
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buckets[frequency] = new ArrayList<>();
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}
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buckets[frequency].add(key);
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}
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List<Integer> topK = new ArrayList<>();
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for (int i = buckets.length - 1; i >= 0 && topK.size() < k; i--) {
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if (buckets[i] == null) {
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continue;
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}
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if (buckets[i].size() <= (k - topK.size())) {
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topK.addAll(buckets[i]);
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} else {
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topK.addAll(buckets[i].subList(0, k - topK.size()));
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}
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}
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return topK;
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public List<Integer> topKFrequent(int[] nums, int k) {
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Map<Integer, Integer> frequencyForNum = new HashMap<>();
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for (int num : nums) {
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frequencyForNum.put(num, frequencyForNum.getOrDefault(num, 0) + 1);
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}
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List<Integer>[] buckets = new ArrayList[nums.length + 1];
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for (int key : frequencyForNum.keySet()) {
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int frequency = frequencyForNum.get(key);
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if (buckets[frequency] == null) {
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buckets[frequency] = new ArrayList<>();
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}
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buckets[frequency].add(key);
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}
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List<Integer> topK = new ArrayList<>();
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for (int i = buckets.length - 1; i >= 0 && topK.size() < k; i--) {
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if (buckets[i] == null) {
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continue;
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}
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if (buckets[i].size() <= (k - topK.size())) {
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topK.addAll(buckets[i]);
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} else {
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topK.addAll(buckets[i].subList(0, k - topK.size()));
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}
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}
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return topK;
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}
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```
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## 按照字符出现次数对字符串排序
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## 按照字符出现次数对字符串排序
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[451. Sort Characters By Frequency (Medium)](https://leetcode.com/problems/sort-characters-by-frequency/description/)
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[451. Sort Characters By Frequency (Medium)](https://leetcode.com/problems/sort-characters-by-frequency/description/)
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```html
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Input:
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@ -150,40 +138,40 @@ Output:
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"eert"
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Explanation:
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'e' appears twice while 'r' and 't' both appear once.
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So 'e' must appear before both 'r' and 't'. Therefore "eetr" is also a valid answer.
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'e' appears twice while 'r' and 't' both appear once.
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So 'e' must appear before both 'r' and 't'. Therefore "eetr" is also a valid answer.
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```
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```java
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public String frequencySort(String s) {
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Map<Character, Integer> frequencyForNum = new HashMap<>();
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for (char c : s.toCharArray())
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frequencyForNum.put(c, frequencyForNum.getOrDefault(c, 0) + 1);
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public String frequencySort(String s) {
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Map<Character, Integer> frequencyForNum = new HashMap<>();
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for (char c : s.toCharArray())
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frequencyForNum.put(c, frequencyForNum.getOrDefault(c, 0) + 1);
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List<Character>[] frequencyBucket = new ArrayList[s.length() + 1];
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for (char c : frequencyForNum.keySet()) {
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int f = frequencyForNum.get(c);
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if (frequencyBucket[f] == null) {
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frequencyBucket[f] = new ArrayList<>();
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}
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frequencyBucket[f].add(c);
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}
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StringBuilder str = new StringBuilder();
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for (int i = frequencyBucket.length - 1; i >= 0; i--) {
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if (frequencyBucket[i] == null) {
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continue;
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}
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for (char c : frequencyBucket[i]) {
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for (int j = 0; j < i; j++) {
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str.append(c);
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}
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}
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}
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return str.toString();
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List<Character>[] frequencyBucket = new ArrayList[s.length() + 1];
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for (char c : frequencyForNum.keySet()) {
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int f = frequencyForNum.get(c);
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if (frequencyBucket[f] == null) {
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frequencyBucket[f] = new ArrayList<>();
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}
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frequencyBucket[f].add(c);
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}
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StringBuilder str = new StringBuilder();
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for (int i = frequencyBucket.length - 1; i >= 0; i--) {
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if (frequencyBucket[i] == null) {
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continue;
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}
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for (char c : frequencyBucket[i]) {
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for (int j = 0; j < i; j++) {
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str.append(c);
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}
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}
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}
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return str.toString();
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}
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```
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# 荷兰国旗问题
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# 荷兰国旗问题
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荷兰国旗包含三种颜色：红、白、蓝。
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@ -191,36 +179,36 @@ public String frequencySort(String s) {
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它其实是三向切分快速排序的一种变种，在三向切分快速排序中，每次切分都将数组分成三个区间：小于切分元素、等于切分元素、大于切分元素，而该算法是将数组分成三个区间：等于红色、等于白色、等于蓝色。
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<div align="center"> <img src="pics/7a3215ec-6fb7-4935-8b0d-cb408208f7cb.png"/> </div><br>
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## 按颜色进行排序
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## 按颜色进行排序
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[75. Sort Colors (Medium)](https://leetcode.com/problems/sort-colors/description/)
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[75. Sort Colors (Medium)](https://leetcode.com/problems/sort-colors/description/)
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```html
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Input: [2,0,2,1,1,0]
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Output: [0,0,1,1,2,2]
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Input: [2,0,2,1,1,0]
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Output: [0,0,1,1,2,2]
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```
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题目描述：只有 0/1/2 三种颜色。
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题目描述：只有 0/1/2 三种颜色。
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```java
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public void sortColors(int[] nums) {
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int zero = -1, one = 0, two = nums.length;
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while (one < two) {
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if (nums[one] == 0) {
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swap(nums, ++zero, one++);
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} else if (nums[one] == 2) {
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swap(nums, --two, one);
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} else {
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++one;
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}
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}
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public void sortColors(int[] nums) {
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int zero = -1, one = 0, two = nums.length;
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while (one < two) {
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if (nums[one] == 0) {
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swap(nums, ++zero, one++);
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} else if (nums[one] == 2) {
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swap(nums, --two, one);
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} else {
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++one;
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}
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}
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}
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private void swap(int[] nums, int i, int j) {
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int t = nums[i];
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nums[i] = nums[j];
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nums[j] = t;
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private void swap(int[] nums, int i, int j) {
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int t = nums[i];
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nums[i] = nums[j];
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nums[j] = t;
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}
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```
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