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notes/Leetcode 题解.md
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@ -331,51 +331,15 @@ You need to output 2.
public int findContentChildren(int[] g, int[] s) { public int findContentChildren(int[] g, int[] s) {
Arrays.sort(g); Arrays.sort(g);
Arrays.sort(s); Arrays.sort(s);
int i = 0, j = 0; int gIndex = 0, sIndex = 0;
while(i < g.length && j < s.length){ while (gIndex < g.length && sIndex < s.length) {
if(g[i] <= s[j]) i++; if (g[gIndex] <= s[sIndex]) gIndex++;
j++; sIndex++;
} }
return i; return gIndex;
} }
``` ```
**投飞镖刺破气球**
[Leetcode : 452. Minimum Number of Arrows to Burst Balloons (Medium)](https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons/description/)
```
Input:
[[10,16], [2,8], [1,6], [7,12]]
Output:
2
```
题目描述：气球在一个水平数轴上摆放，可以重叠，飞镖垂直射向坐标轴，使得路径上的气球都会刺破。求解最小的投飞镖次数使所有气球都被刺破。
从左往右投飞镖，并且在每次投飞镖时满足以下条件：
1. 左边已经没有气球了；
2. 本次投飞镖能够刺破最多的气球。
```java
public int findMinArrowShots(int[][] points) {
if(points.length == 0) return 0;
Arrays.sort(points,(a,b) -> (a[1] - b[1]));
int curPos = points[0][1];
int ret = 1;
for (int i = 1; i < points.length; i++) {
if(points[i][0] <= curPos) {
continue;
}
curPos = points[i][1];
ret++;
}
return ret;
}
```
**股票的最大收益** **股票的最大收益**
[Leetcode : 122. Best Time to Buy and Sell Stock II (Easy)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/description/) [Leetcode : 122. Best Time to Buy and Sell Stock II (Easy)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/description/)
@ -387,8 +351,10 @@ public int findMinArrowShots(int[][] points) {
```java ```java
public int maxProfit(int[] prices) { public int maxProfit(int[] prices) {
int profit = 0; int profit = 0;
for(int i = 1; i < prices.length; i++){ for (int i = 1; i < prices.length; i++) {
if(prices[i] > prices[i-1]) profit += (prices[i] - prices[i-1]); if (prices[i] > prices[i - 1]) {
profit += (prices[i] - prices[i - 1]);
}
} }
return profit; return profit;
} }
@ -403,16 +369,16 @@ Input: flowerbed = [1,0,0,0,1], n = 1
Output: True Output: True
``` ```
题目描述：花朵之间至少需要一个单位的间隔。 题目描述：花朵之间至少需要一个单位的间隔，求解是否能种下 n 朵花。
```java ```java
public boolean canPlaceFlowers(int[] flowerbed, int n) { public boolean canPlaceFlowers(int[] flowerbed, int n) {
int cnt = 0; int cnt = 0;
for(int i = 0; i < flowerbed.length; i++){ for (int i = 0; i < flowerbed.length; i++) {
if(flowerbed[i] == 1) continue; if (flowerbed[i] == 1) continue;
int pre = i == 0 ? 0 : flowerbed[i - 1]; int pre = i == 0 ? 0 : flowerbed[i - 1];
int next = i == flowerbed.length - 1 ? 0 : flowerbed[i + 1]; int next = i == flowerbed.length - 1 ? 0 : flowerbed[i + 1];
if(pre == 0 && next == 0) { if (pre == 0 && next == 0) {
cnt++; cnt++;
flowerbed[i] = 1; flowerbed[i] = 1;
} }
@ -433,15 +399,15 @@ Explanation: You could modify the first 4 to 1 to get a non-decreasing array.
题目描述：判断一个数组能不能只修改一个数就成为非递减数组。 题目描述：判断一个数组能不能只修改一个数就成为非递减数组。
在出现 nums[i] < nums[i - 1] 时，需要考虑的是应该修改数组的哪个数，使得本次修改能使 i 之前的数组成为非递减数组，并且 **不影响后续的操作** 。优先考虑令 nums[i - 1] = nums[i]，因为如果修改 nums[i] = nums[i - 1] 的话，那么 nums[i] 这个数会变大，那么就有可能比 nums[i + 1] 大，从而影响了后续操作。还有一个比较特别的情况就是 nums[i] < nums[i - 2]，只修改 nums[i - 1] = nums[i] 不能令数组成为非递减，只能通过修改 nums[i] = nums[i - 1] 才行。 在出现 nums[i] < nums[i - 1] 时，需要考虑的是应该修改数组的哪个数，使得本次修改能使 i 之前的数组成为非递减数组，并且 **不影响后续的操作** 。优先考虑令 nums[i - 1] = nums[i]，因为如果修改 nums[i] = nums[i - 1] 的话，那么 nums[i] 这个数会变大，就有可能比 nums[i + 1] 大，从而影响了后续操作。还有一个比较特别的情况就是 nums[i] < nums[i - 2]，只修改 nums[i - 1] = nums[i] 不能令数组成为非递减，只能通过修改 nums[i] = nums[i - 1] 才行。
```java ```java
public boolean checkPossibility(int[] nums) { public boolean checkPossibility(int[] nums) {
int cnt = 0; int cnt = 0;
for(int i = 1; i < nums.length; i++){ for (int i = 1; i < nums.length; i++) {
if(nums[i] < nums[i - 1]){ if (nums[i] < nums[i - 1]) {
cnt++; cnt++;
if(i - 2 >= 0 && nums[i - 2] > nums[i]) nums[i] = nums[i-1]; if (i - 2 >= 0 && nums[i - 2] > nums[i]) nums[i] = nums[i - 1];
else nums[i - 1] = nums[i]; else nums[i - 1] = nums[i];
} }
} }
@ -460,15 +426,54 @@ Return true.
```java ```java
public boolean isSubsequence(String s, String t) { public boolean isSubsequence(String s, String t) {
int index = 0; int pos = -1;
for (char c : s.toCharArray()) { for (char c : s.toCharArray()) {
index = t.indexOf(c, index); pos = t.indexOf(c, pos + 1);
if (index == -1) return false; if (pos == -1) return false;
} }
return true; return true;
} }
``` ```
**投飞镖刺破气球**
[Leetcode : 452. Minimum Number of Arrows to Burst Balloons (Medium)](https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons/description/)
```
Input:
[[10,16], [2,8], [1,6], [7,12]]
Output:
2
```
题目描述：气球在一个水平数轴上摆放，可以重叠，飞镖垂直投向坐标轴，使得路径上的气球都会刺破。求解最小的投飞镖次数使所有气球都被刺破。
对气球按末尾位置进行排序，得到：
```html
[[1,6], [2,8], [7,12], [10,16]]
```
如果让飞镖投向 6 这个位置，那么 [1,6] 和 [2,8] 这两个气球都会被刺破，这种方式下刺破这两个气球的投飞镖次数最少，并且后面两个气球依然可以使用这种方式来刺破。
```java
public int findMinArrowShots(int[][] points) {
if (points.length == 0) return 0;
Arrays.sort(points, (a, b) -> (a[1] - b[1]));
int curPos = points[0][1];
int shots = 1;
for (int i = 1; i < points.length; i++) {
if (points[i][0] <= curPos) {
continue;
}
curPos = points[i][1];
shots++;
}
return shots;
}
```
**分隔字符串使同种字符出现在一起** **分隔字符串使同种字符出现在一起**
[Leetcode : 763. Partition Labels (Medium)](https://leetcode.com/problems/partition-labels/description/) [Leetcode : 763. Partition Labels (Medium)](https://leetcode.com/problems/partition-labels/description/)