★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：https://www.cnblogs.com/strengthen/p/9791084.html 
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

热烈欢迎，请直接点击！！！

进入博主App Store主页，下载使用各个作品！！！

注：博主将坚持每月上线一个新app！！！

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.

假设你是一位很棒的家长，想要给你的孩子们一些小饼干。但是，每个孩子最多只能给一块饼干。对每个孩子 i ，都有一个胃口值 gi ，这是能让孩子们满足胃口的饼干的最小尺寸；并且每块饼干 j ，都有一个尺寸 sj 。如果 sj >= gi ，我们可以将这个饼干 j 分配给孩子 i ，这个孩子会得到满足。你的目标是尽可能满足越多数量的孩子，并输出这个最大数值。

注意：

你可以假设胃口值为正。
一个小朋友最多只能拥有一块饼干。

示例 1:

输入: [1,2,3], [1,1]

输出: 1

解释: 
你有三个孩子和两块小饼干，3个孩子的胃口值分别是：1,2,3。
虽然你有两块小饼干，由于他们的尺寸都是1，你只能让胃口值是1的孩子满足。
所以你应该输出1。

示例 2:

输入: [1,2], [1,2,3]

输出: 2

解释: 
你有两个孩子和三块小饼干，2个孩子的胃口值分别是1,2。
你拥有的饼干数量和尺寸都足以让所有孩子满足。
所以你应该输出2.

64ms

 1 class Solution {
 2     func findContentChildren(_ g: [Int], _ s: [Int]) -> Int {
 3         //对数组进行排序
 4         var sortedS = s.sorted(by:<)
 5         var sortedG = g.sorted(by:<)
 6         var i=0, j=0
 7         while i < s.count && j < g.count {
 8             if sortedS[i] >= sortedG[j] {
 9                 j+=1
10             }
11             i+=1
12         }
13         return j
14     }
15 }

108ms

 1 class Solution {
 2     func findContentChildren(_ g: [Int], _ s: [Int]) -> Int {
 3         var count = 0
 4     var childs = g.sorted(by: >)
 5     var cookies = s.sorted(by: >)
 6     
 7     for i in 0..<childs.count {
 8         if let bc = cookies.first{
 9             if bc >= childs[i] {
10                 count += 1
11                 cookies.removeFirst()
12             }
13         }
14         
15     }
16     
17     return count
18     }
19 }

120ms

 1 class Solution {
 2     func findContentChildren(_ g: [Int], _ s: [Int]) -> Int {
 3         var sortedS = s.sorted()
 4         var sortedG = g.sorted()
 5         var count = 0
 6         var i=0, j=0
 7         while i < s.count && j < g.count {
 8             if sortedS[i] >= sortedG[j] {
 9                 j+=1
10             }
11             i+=1
12         }
13         return j
14     }
15 }