题目描述：
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.
要完成的函数：
int findContentChildren(vector<int>& g, vector<int>& s)
说明：
1、这道题目看清题意就不难了。每个孩子有个需求尺寸数，是整数，1、2、3这样子的。有几块饼干，尺寸是整数，1、2、3这样子的。不能把饼干掰开给。
要求输出能够满足几个孩子的需求。
2、首先对需求尺寸数g进行排序，对饼干尺寸数s进行排序，然后从g和s的第一位开始比较，直到g或者s处理完了。
代码也很简单，如下：
int findContentChildren(vector<int>& g, vector<int>& s)
{
sort(g.begin(),g.end());
sort(s.begin(),s.end());
int i=0,j=0;
while(i<g.size()&&j<s.size())//处理完所有孩子的需求或者已经没有满足需求的饼干了
{
if(s[j]>=g[i])
i++; //如果满足了处理下一个i
j++;//如果不满足j一直加
}
return i;
}
上述代码实测40ms，beats 97.41% of cpp submissions。