(LeetCode 561-Easy)Array Partition I

June 13th, 2017

https://leetcode.com/problems/array-partition-i

Description

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].

问题分析：

由题意可知，我们可以将整个数组升序排序，每两个数为一组，取其最小的那个值相加，最终得到答案。

javascript:

/**
 * @param {number[]} nums
 * @return {number}
 */
var arrayPairSum = function(nums) {
    var len = nums.length,
    total = nums.sort((x,y)=>x-y),
    sum=0;
    
    do {
        sum +=total[len-2];
    } while (len-=2);
 
    return sum;
    
};

python:

class Solution(object):

    def arrayPairSum(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        return sum(sorted(nums)[::2])


class Solution(object):
    def arrayPairSum(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        
        nums.sort()
        return sum(nums[::2])

Demi_YuHongJun

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