## 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.``````

## 解法一

### Python

``````class Solution:
def findContentChildren(self, g, s):
"""
:type g: List[int]
:type s: List[int]
:rtype: int
"""
g = sorted(g)
s = sorted(s)
i = 0
count = 0
for item in g:
while i < len(s) and item > s[i]:
i += 1
if i == len(s):
break
count += 1
i += 1
return count``````

### Java

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

### C++

``````class Solution {
public:
int findContentChildren(vector<int> &g, vector<int> &s) {
sort(g.begin(), g.end());
sort(s.begin(), s.end());
int i = 0;
int count = 0;
for (auto item : g) {
while (i < s.size() && item > s[i]) {
i++;
}
if (i == s.size()) {
break;
}
count++;
i++;
}
return count;
}
};``````