题目 #

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.

解题思路 #

• 假设你想给小朋友们饼干，每个小朋友最多能够给一块饼干。每个小朋友都有一个“贪心指数”，称为 g[i]g[i] 表示的是这名小朋友需要的饼干大小的最小值。同时，每个饼干都有一个大小值 s[i]，如果 s[j] ≥ g[i]，我们将饼干 j 分给小朋友 i 后，小朋友会很开心。给定数组 g[]s[]，问如何分配饼干，能让更多的小朋友开心。
• 这是一道典型的简单贪心题。贪心题一般都伴随着排序。将 g[]s[] 分别排序。按照最难满足的小朋友开始给饼干，依次往下满足，最终能满足的小朋友数就是最终解。

代码 #

package leetcode

import "sort"

func findContentChildren(g []int, s []int) int {
sort.Ints(g)
sort.Ints(s)
gi, si, res := 0, 0, 0
for gi < len(g) && si < len(s) {
if s[si] >= g[gi] {
res++
si++
gi++
} else {
si++
}
}
return res
}

Apr 8, 2023