####QUESTION 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.

####EXPLANATION
其实很简单的，首先将两个数组排序，然后将第一个cookies进行分配，找到满足的之后，进行第二个cookies的分配。

但是这个解法其实有一个问题：如果数组是【2，3】【1，1，1，1】的话，其实输出应该是0，但是还是需要进行m*n次的操作。
该解法的时间复杂度是：O(m*n)并不能算是最优的解法。
####SOLUTION

```
public int findContentChildren(int[] g, int[] s) {
int count = 0;
Arrays.sort(g);
Arrays.sort(s);
for(int i = 0 ; i<s.length;i++){
for(int j = count ;j<g.length;j++){
if(s[i]>=g[j]){
count++;
break;
}
}
}
return count;
}
```