LEETCODE--Assign Cookies

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.

原创算法（runtime略靠后）：
先将greed，size两个vector中的数据排序，然后再从greed前端开始依次从size-vector中（从大到小）去寻找“最小满足的size”，直至greed中有一个值无法从整个size-vector中找到合适的size.则此greed以后的所有
greed全都不可能找到满足的size了。从而可以得到合适的人数。

class Solution {public:    int findContentChildren(vector<int>& g, vector<int>& s) {        sort(g.begin(), g.end());        sort(s.begin(), s.end());        int count = 0;        int glen = g.size();        int slen = s.size();        int j = 0;        for(; count < glen; count++){            int tag = 0;            for(; j < slen; j++){                if(g[count] <= s[j]){                    tag = 1;                    j++;                    break;                }            }            if(tag == 0)                break;        }        return count;    }};

方法二：
用一个while循环解决问题

class Solution {public:    int findContentChildren(vector<int>& g, vector<int>& s) {        sort(g.begin(), g.end());        sort(s.begin(), s.end());        int glen = g.size();        int slen = s.size();        int count = 0;        int j = 0;        while(count < glen && j < slen){            if(g[count] <= s[j])                count++;            j++;        }        return count;    }};