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: 1 Explanation: 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: 2 Explanation: 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.
题解：
先把两个vector升序排列，这样再从头遍历，当cookie遍历完就结束。要注意遍历到cookie最后一项时，满足要求和不满足要求两种情况的处理。还有，当g[i]<=s[k]时，如果直接break，k并没有+1，就会导致重复遍历了一个cookie。
答案：
class Solution { public: int findContentChildren(vector<int>& g, vector<int>& s) { sort(g.begin(),g.end()); sort(s.begin(),s.end()); int result=0; int k=0; bool even=true; for(int i=0;i<g.size();i++) { for(;k<s.size();k++) { if(g[i]<=s[k]) { result++; if(k!=s.size()-1) k++; else even=false; break; } else if(k==s.size()-1) { even=false; break; } } if(!even) break; } return result; } };
