题目地址:https://leetcode.com/problems/assign-cookies/open in new window

  • 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 jto 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、 Youmayassumethegreedfactorisalwayspositive.;
2、 Youcannotassignmorethanonecookietoonechild.;

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.

题目大意

我们需要把饼干分给一群小孩,每个小孩最多只有一个饼干。每个小孩有自己的欲望,每个饼干有自己的大小,给小孩分饼干的时候最少也要给他欲望大小的饼干。求能让多少个小孩满意。

解题方法

明显的贪心算法,尽可能给小孩满足他欲望的最小饼干。

Java解法

这个题目简单的思路就是:

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;
    }
}

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AC:17ms

Python解法

二刷,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

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