zoukankan      html  css  js  c++  java
  • 561. Array Partition I (Easy)

    Given an array of 2n integers, your task is to group these integers into n pairs of integer, say (a1, b1), (a2, b2), ..., (an, bn) which makes sum of min(ai, bi) for all i from 1 to n as large as possible.

    Example 1:

    Input: [1,4,3,2]
    
    Output: 4
    Explanation: n is 2, and the maximum sum of pairs is 4 = min(1, 2) + min(3, 4).
    

    Note:

    1. n is a positive integer, which is in the range of [1, 10000].
    2. All the integers in the array will be in the range of [-10000, 10000].

    思路:排序(sort)

    1.求每两个数之间的最小值,所有值和的最大值;
    2.数组从小到大排序(使每一对的两个数字尽可能相近);

    class Solution():
        def arrayPairSum(self, nums):
            """
            :type nums: List[int]
            :rtype: int
            """
            return sum(sorted(nums)[::2])
  • 相关阅读:
    c++类的知识点(1)
    并查集经典例题分析
    并查集
    bfs-迷宫
    出栈次序--数学归纳法--蓝桥
    九宫重排
    Tomcat详解
    寒假日记-第三天
    寒假日记-第二天(MySQL语句)
    Java学期总结
  • 原文地址:https://www.cnblogs.com/yancea/p/7507592.html
Copyright © 2011-2022 走看看