LeetCode 455 Assign Cookies

题目：

Assume you are an awesome parent and want to give your children some cookies. But, you should give each child at most one cookie. Each child i has a greed factor gi, which is the minimum size of a cookie that the child will be content with; and each cookie j has a size sj. If sj >= gi, we can assign the cookie j to the child i, and the child i will be content. Your goal is to maximize the number of your content children and output the maximum number.

Note:
You may assume the greed factor is always positive. 
You cannot assign more than one cookie to one child.

Example 1:

Input: [1,2,3], [1,1]Output: 1Explanation: You have 3 children and 2 cookies. The greed factors of 3 children are 1, 2, 3. And even though you have 2 cookies, since their size is both 1, you could only make the child whose greed factor is 1 content.You need to output 1.

Example 2:

Input: [1,2], [1,2,3]Output: 2Explanation: You have 2 children and 3 cookies. The greed factors of 2 children are 1, 2. You have 3 cookies and their sizes are big enough to gratify all of the children, You need to output 2.

题目链接

题意：

题目具体的意思就不多描述了，翻译即可，我理解的是给两个数组 g 和 s 。问 s 中有最多有多少个元素可以与 g 中的元素一一映射且 大于等于g中与其一一映射的的元素？

思路是先降序排序，之后枚举数组s，对于每一个s元素，寻找g中从前一个的坐标开始，第一个小于它的元素的。以此类推

代码如下：

class Solution {public:    int findContentChildren(vector<int>& g, vector<int>& s) {        sort(g.begin(), g.end(), greater<int>());        sort(s.begin(), s.end(), greater<int>());        int ans = 0, greed = 0, size = 0;        for (; size < s.size(); size ++, greed ++) {            while (greed < g.size() && g[greed] > s[size]) greed ++;            if (greed < g.size() && g[greed] <= s[size])                ans ++;        }        return ans;    }};