From cf796faa06681b47cd17f62e948fd0ef471d172d Mon Sep 17 00:00:00 2001 From: CyC2018 <1029579233@qq.com> Date: Wed, 4 Apr 2018 15:36:04 +0800 Subject: [PATCH] auto commit --- notes/Leetcode 题解.md | 172 ++++++++++++++++++--------------------- notes/剑指 offer 题解.md | 2 + 2 files changed, 81 insertions(+), 93 deletions(-) diff --git a/notes/Leetcode 题解.md b/notes/Leetcode 题解.md index 53292384..dc22b4b3 100644 --- a/notes/Leetcode 题解.md +++ b/notes/Leetcode 题解.md @@ -3223,16 +3223,16 @@ public int[] countBits(int num) { 一个栈实现： ```java -class MyQueue { +class MyQueue { private Stack st = new Stack(); public void push(int x) { Stack temp = new Stack(); - while(!st.isEmpty()){ + while (!st.isEmpty()) { temp.push(st.pop()); } st.push(x); - while(!temp.isEmpty()){ + while (!temp.isEmpty()) { st.push(temp.pop()); } } @@ -3254,7 +3254,7 @@ class MyQueue { 两个栈实现： ```java -class MyQueue { +class MyQueue { private Stack in = new Stack(); private Stack out = new Stack(); @@ -3272,9 +3272,9 @@ class MyQueue { return out.peek(); } - private void in2out(){ - if(out.isEmpty()){ - while(!in.isEmpty()){ + private void in2out() { + if (out.isEmpty()) { + while (!in.isEmpty()) { out.push(in.pop()); } } @@ -3301,8 +3301,9 @@ class MyStack { public void push(int x) { queue.add(x); - for(int i = 1; i < queue.size(); i++){ // 翻转 - queue.add(queue.remove()); + int cnt = queue.size(); + while (cnt-- > 1) { + queue.add(queue.poll()); } } @@ -3341,20 +3342,14 @@ class MinStack { public void push(int x) { dataStack.add(x); - if(x < min) { - min = x; - } + min = Math.min(min, x); minStack.add(min); } public void pop() { dataStack.pop(); minStack.pop(); - if(!minStack.isEmpty()) { - min = minStack.peek(); - } else{ - min = Integer.MAX_VALUE; - } + min = minStack.isEmpty() ? min = Integer.MAX_VALUE : minStack.peek(); } public int top() { @@ -3382,17 +3377,15 @@ Output : true ```java public boolean isValid(String s) { Stack stack = new Stack<>(); - for(int i = 0; i < s.length(); i++){ - char c = s.charAt(i); - if(c == '(' || c == '{' || c == '[') stack.push(c); - else{ - if(stack.isEmpty()) return false; + for (char c : s.toCharArray()) { + if (c == '(' || c == '{' || c == '[') stack.push(c); + else { + if (stack.isEmpty()) return false; char cStack = stack.pop(); - if(c == ')' && cStack != '(' || - c == ']' && cStack != '[' || - c == '}' && cStack != '{' ) { - return false; - } + boolean b1 = c == ')' && cStack != '('; + boolean b2 = c == ']' && cStack != '['; + boolean b3 = c == '}' && cStack != '{'; + if (b1 || b2 || b3) return false; } } return stack.isEmpty(); @@ -3729,14 +3722,9 @@ For example, given nums = [0, 1, 0, 3, 12], after calling your function, nums sh ```java public void moveZeroes(int[] nums) { - int n = nums.length; int idx = 0; - for(int i = 0; i < n; i++){ - if(nums[i] != 0) nums[idx++] = nums[i]; - } - while(idx < n){ - nums[idx++] = 0; - } + for (int num : nums) if (num != 0) nums[idx++] = num; + while (idx < nums.length) nums[idx++] = 0; } ``` @@ -3751,6 +3739,11 @@ Input: nums = [1,2,2,4] Output: [2,3] ``` +```html +Input: nums = [1,2,2,4] +Output: [2,3] +``` + 最直接的方法是先对数组进行排序，这种方法时间复杂度为 O(nlogn)。本题可以以 O(n) 的时间复杂度、O(1) 空间复杂度来求解。 主要思想是通过交换数组元素，使得数组上的元素在正确的位置上。 @@ -3764,32 +3757,29 @@ Output: [2,3] ```java public int[] findErrorNums(int[] nums) { - for(int i = 0; i < nums.length; i++){ - while(nums[i] != i + 1 && nums[i] != nums[nums[i] - 1]) { + for (int i = 0; i < nums.length; i++) { + while (nums[i] != i + 1) { + if (nums[i] == nums[nums[i] - 1]) { + return new int[]{nums[nums[i] - 1], i + 1}; + } swap(nums, i, nums[i] - 1); } } - for(int i = 0; i < nums.length; i++){ - if(i + 1 != nums[i]) { - return new int[]{nums[i], i + 1}; - } - } - return null; } -private void swap(int[] nums, int i, int j){ - int tmp = nums[i]; - nums[i] = nums[j]; - nums[j] = tmp; +private void swap(int[] nums, int i, int j) { + int tmp = nums[i]; nums[i] = nums[j]; nums[j] = tmp; } ``` -**找出数组中重复的数，数组值在 [0, n-1] 之间** +**找出数组中重复的数，数组值在 [1, n] 之间** [Leetcode : 287. Find the Duplicate Number (Medium)](https://leetcode.com/problems/find-the-duplicate-number/description/) +要求不能修改数组，也不能使用额外的空间。 + 二分查找解法： ```java @@ -3812,18 +3802,17 @@ public int findDuplicate(int[] nums) { ```java public int findDuplicate(int[] nums) { - int slow = nums[0], fast = nums[nums[0]]; - while (slow != fast) { - slow = nums[slow]; - fast = nums[nums[fast]]; - } - - fast = 0; - while (slow != fast) { - slow = nums[slow]; - fast = nums[fast]; - } - return slow; + int slow = nums[0], fast = nums[nums[0]]; + while (slow != fast) { + slow = nums[slow]; + fast = nums[nums[fast]]; + } + fast = 0; + while (slow != fast) { + slow = nums[slow]; + fast = nums[fast]; + } + return slow; } ``` @@ -4014,18 +4003,20 @@ public ListNode deleteDuplicates(ListNode head) { [Leetcode : 234. Palindrome Linked List (Easy)](https://leetcode.com/problems/palindrome-linked-list/description/) +要求以 O(1) 的空间复杂度来求解。 + 切成两半，把后半段反转，然后比较两半是否相等。 ```java public boolean isPalindrome(ListNode head) { - if(head == null || head.next == null) return true; + if (head == null || head.next == null) return true; ListNode slow = head, fast = head.next; - while(fast != null && fast.next != null){ + while (fast != null && fast.next != null) { slow = slow.next; fast = fast.next.next; } - if(fast != null){ // 偶数节点，让 slow 指向下一个节点 + if (fast != null) { // 偶数节点，让 slow 指向下一个节点 slow = slow.next; } @@ -4035,14 +4026,14 @@ public boolean isPalindrome(ListNode head) { return isEqual(l1, l2); } -private void cut(ListNode head, ListNode cutNode){ - while( head.next != cutNode ) head = head.next; +private void cut(ListNode head, ListNode cutNode) { + while (head.next != cutNode) head = head.next; head.next = null; } -private ListNode reverse(ListNode head){ +private ListNode reverse(ListNode head) { ListNode newHead = null; - while(head != null){ + while (head != null) { ListNode nextNode = head.next; head.next = newHead; newHead = head; @@ -4051,9 +4042,9 @@ private ListNode reverse(ListNode head){ return newHead; } -private boolean isEqual(ListNode l1, ListNode l2){ - while(l1 != null && l2 != null){ - if(l1.val != l2.val) return false; +private boolean isEqual(ListNode l1, ListNode l2) { + while (l1 != null && l2 != null) { + if (l1.val != l2.val) return false; l1 = l1.next; l2 = l2.next; } @@ -4307,19 +4298,13 @@ There are two left leaves in the binary tree, with values 9 and 15 respectively. ```java public int sumOfLeftLeaves(TreeNode root) { - if(root == null) { - return 0; - } - if(isLeaf(root.left)) { - return root.left.val + sumOfLeftLeaves(root.right); - } + if(root == null) return 0; + if(isLeaf(root.left)) return root.left.val + sumOfLeftLeaves(root.right); return sumOfLeftLeaves(root.left) + sumOfLeftLeaves(root.right); } private boolean isLeaf(TreeNode node){ - if(node == null) { - return false; - } + if(node == null) return false; return node.left == null && node.right == null; } ``` @@ -4376,7 +4361,7 @@ Given tree s: / \ 1 2 Given tree t: - 4 + 4 / \ 1 2 Return true, because t has the same structure and node values with a subtree of s. @@ -4651,11 +4636,11 @@ public int findBottomLeftValue(TreeNode root) { ### 前中后序遍历 ```html - 1 - / \ - 2 3 - / \ \ -4 5 6 + 1 + / \ + 2 3 + / \ \ +4 5 6 ``` 层次遍历顺序：[1 2 3 4 5 6] @@ -4704,14 +4689,14 @@ void dfs(TreeNode root){ ```java public List preorderTraversal(TreeNode root) { List ret = new ArrayList<>(); - if (root == null) return ret; Stack stack = new Stack<>(); stack.push(root); while (!stack.isEmpty()) { TreeNode node = stack.pop(); + if (node == null) continue; ret.add(node.val); - if (node.right != null) stack.push(node.right); - if (node.left != null) stack.push(node.left); // 先添加右子树再添加左子树，这样是为了让左子树在栈顶 + stack.push(node.right); // 先右后左，保证左子树先遍历 + stack.push(node.left); } return ret; } @@ -4726,14 +4711,14 @@ public List preorderTraversal(TreeNode root) { ```java public List postorderTraversal(TreeNode root) { List ret = new ArrayList<>(); - if (root == null) return ret; Stack stack = new Stack<>(); stack.push(root); while (!stack.isEmpty()) { TreeNode node = stack.pop(); + if (node == null) continue; ret.add(node.val); - if (node.left != null) stack.push(node.left); - if (node.right != null) stack.push(node.right); + stack.push(node.left); + stack.push(node.right); } Collections.reverse(ret); return ret; @@ -4747,11 +4732,12 @@ public List postorderTraversal(TreeNode root) { ```java public List inorderTraversal(TreeNode root) { List ret = new ArrayList<>(); + if (root == null) return ret; Stack stack = new Stack<>(); TreeNode cur = root; - while(cur != null || !stack.isEmpty()) { - while(cur != null) { // 模拟递归栈的不断深入 - stack.add(cur); + while (cur != null || !stack.isEmpty()) { + while (cur != null) { + stack.push(cur); cur = cur.left; } TreeNode node = stack.pop(); diff --git a/notes/剑指 offer 题解.md b/notes/剑指 offer 题解.md index cdc4e4c1..2dc15d12 100644 --- a/notes/剑指 offer 题解.md +++ b/notes/剑指 offer 题解.md @@ -90,6 +90,8 @@ 在一个长度为 n 的数组里的所有数字都在 0 到 n-1 的范围内。数组中某些数字是重复的，但不知道有几个数字是重复的。也不知道每个数字重复几次。请找出数组中任意一个重复的数字。例如，如果输入长度为 7 的数组 {2, 3, 1, 0, 2, 5, 3}，那么对应的输出是第一个重复的数字 2。 +要求复杂度为 O(N) + O(1)，时间复杂度 O(N)，空间复杂度 O(1)。因此不能使用排序的方法，也不能使用额外的标记数组。 + ## 解题思路 这种数组元素在 [0, n-1] 范围内的问题，可以将值为 i 的元素放到第 i 个位置上。