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CyC2018
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@ -331,48 +331,12 @@ You need to output 2.
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public int findContentChildren(int[] g, int[] s) {
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Arrays.sort(g);
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Arrays.sort(s);
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int i = 0, j = 0;
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while(i < g.length && j < s.length){
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if(g[i] <= s[j]) i++;
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j++;
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int gIndex = 0, sIndex = 0;
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while (gIndex < g.length && sIndex < s.length) {
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if (g[gIndex] <= s[sIndex]) gIndex++;
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sIndex++;
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}
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return i;
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}
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```
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**投飞镖刺破气球**
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[Leetcode : 452. Minimum Number of Arrows to Burst Balloons (Medium)](https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons/description/)
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```
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Input:
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[[10,16], [2,8], [1,6], [7,12]]
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Output:
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2
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```
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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。
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从左往右投飞镖，并且在每次投飞镖时满足以下条件：
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1. 左边已经没有气球了；
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2. 本次投飞镖能够刺破最多的气球。
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```java
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public int findMinArrowShots(int[][] points) {
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if(points.length == 0) return 0;
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Arrays.sort(points,(a,b) -> (a[1] - b[1]));
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int curPos = points[0][1];
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int ret = 1;
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for (int i = 1; i < points.length; i++) {
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if(points[i][0] <= curPos) {
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continue;
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}
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curPos = points[i][1];
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ret++;
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}
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return ret;
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return gIndex;
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}
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```
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@ -388,7 +352,9 @@ public int findMinArrowShots(int[][] points) {
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public int maxProfit(int[] prices) {
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int profit = 0;
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for (int i = 1; i < prices.length; i++) {
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if(prices[i] > prices[i-1]) profit += (prices[i] - prices[i-1]);
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if (prices[i] > prices[i - 1]) {
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profit += (prices[i] - prices[i - 1]);
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}
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}
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return profit;
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}
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@ -403,7 +369,7 @@ Input: flowerbed = [1,0,0,0,1], n = 1
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Output: True
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```
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题目描述：花朵之间至少需要一个单位的间隔。
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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。
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```java
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public boolean canPlaceFlowers(int[] flowerbed, int n) {
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@ -433,7 +399,7 @@ Explanation: You could modify the first 4 to 1 to get a non-decreasing array.
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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。
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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。
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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。
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```java
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public boolean checkPossibility(int[] nums) {
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@ -460,15 +426,54 @@ Return true.
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```java
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public boolean isSubsequence(String s, String t) {
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int index = 0;
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int pos = -1;
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for (char c : s.toCharArray()) {
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index = t.indexOf(c, index);
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if (index == -1) return false;
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pos = t.indexOf(c, pos + 1);
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if (pos == -1) return false;
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}
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return true;
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}
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```
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**投飞镖刺破气球**
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[Leetcode : 452. Minimum Number of Arrows to Burst Balloons (Medium)](https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons/description/)
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```
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Input:
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[[10,16], [2,8], [1,6], [7,12]]
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Output:
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2
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```
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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。
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对气球按末尾位置进行排序，得到：
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```html
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[[1,6], [2,8], [7,12], [10,16]]
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```
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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。
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```java
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public int findMinArrowShots(int[][] points) {
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if (points.length == 0) return 0;
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Arrays.sort(points, (a, b) -> (a[1] - b[1]));
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int curPos = points[0][1];
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int shots = 1;
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for (int i = 1; i < points.length; i++) {
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if (points[i][0] <= curPos) {
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continue;
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}
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curPos = points[i][1];
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shots++;
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}
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return shots;
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}
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```
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**分隔字符串使同种字符出现在一起**
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[Leetcode : 763. Partition Labels (Medium)](https://leetcode.com/problems/partition-labels/description/)
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