[Swift]LeetCode455. 分发饼干 | Assign Cookies
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➤微信公众号：山青咏芝（shanqingyongzhi）
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➤原文地址：https://www.cnblogs.com/strengthen/p/9791084.html
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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 }