@TOC

[LeetCode]

• Difficulty: Easy

## # 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:

1. You may assume the greed factor is always positive.
2. 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.
``````

## # 解题方法

### # Java解法

1、首先把两个数组排序 2、如果当前满足感小于等于饼干，两个指针都后移，否则，只有满足感后移，然后再和当期前的满足感比较 3、最后返回指向孩子满足感指针的指针位置

``````public class Solution {
public int findContentChildren(int[] g, int[] s) {
Arrays.sort(g);
Arrays.sort(s);
int i=0, j=0;
while(i<g.length && j<s.length){
if(g[i]<=s[j])
i++;
j++;
}
return i;
}
}
``````

AC:17ms

### # Python解法

``````class Solution:
def findContentChildren(self, g, s):
"""
:type g: List[int]
:type s: List[int]
:rtype: int
"""
g.sort()
s.sort()
sp = 0
res = 0
for gi in g:
while sp < len(s) and s[sp] < gi:
sp += 1
if sp < len(s) and s[sp] >= gi:
res += 1
sp += 1
return res
``````

## # 日期

2017 年 1 月 7 日 2018 年 11 月 16 日 —— 又到周五了！