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.
题目标签：Greedy
这道题目给了我们两组array，一组代表小朋友的胃口值，另一组代表曲奇饼干的大小。我们要分发曲奇给尽可能多的小朋友，并且曲奇饼干的大小是可以满足小朋友的胃口的。
所以我们不能把大的曲奇去满足很小胃口的小朋友，除非没有选择。尽可能的去把小曲奇发给小胃口的小朋友。首先把两个array 重新排列，从小到大。然后遍历曲奇array，找到可以满足的小朋友，发给他曲奇，记住他的index，下次就直接从下一个小朋友开始找。
Java Solution:
Runtime beats 57.78%
完成日期：06/24/2017
关键词：Greedy
关键点：把array重新排序
public class Solution
{
public int findContentChildren(int[] g, int[] s)
{
int res = 0;
int index_g = 0;
Arrays.sort(g);
Arrays.sort(s); for(int i=0; i< s.length; i++)
{
if(s[i] >= g[index_g])
{
res++;
index_g++; if(index_g >= g.length)
break;
}
} return res;
}
}
参考资料：N / A
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