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题目地址：

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

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$g_i$, which is the minimum size of a cookie that the child will be content with; and each cookie $j$ has a size $s_j$. If $s_j$ >= $g_i$, we can assign the cookie $j$ to the child $i$, and the child i <script type="math/tex" id="MathJax-Element-684">i</script> 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.

题目比较简单，其实给的示例中已经对我们有提示了，给了两个数组，首先对两个数组进行排序，也就是说对每个孩子的贪婪因子，与蛋糕的大小按从小到大的顺序排序。

首先肯定是看最小的蛋糕能满足谁，然后次小的蛋糕能满足最小贪婪因子后面的哪一位，以此类推：

Java代码实现：

public int findContentChildren(int[] g, int[] s) {    ArrayList
   
     lg = new ArrayList
    
     ();    for (int i = 0; i < g.length; i++) {        lg.add(g[i]);    }    ArrayList
     
       ls = new ArrayList
      
       ();    for (int i = 0; i < s.length; i++) {        ls.add(s[i]);    }    Collections.sort(lg);    Collections.sort(ls);    int content = 0;    int k = 0;    for (int i = 0; i < ls.size(); i++) {        for (int j = k; j < lg.size(); j++) {            if (ls.get(i) < lg.get(j))                continue;            else {                content += 1;                k = j + 1;                break;            }        }    }    return content;}
      
     
    
   

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