[LeetCode] Array Partition I

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].

给定一个含有2n个整数的数组，将这2n个整数分成n个整数对，对每个整数对中的较小数求和，并使这个和最大。根据题意，要使总和最大，在整个数组中需要每个整数对中较小数的取值尽可能大。先将数组排序，此时把两两相邻的整数作为一个整数对，求每个整数对较小数的和，即求排序后数组奇数次序的数值之和。

class Solution {
public:
    int arrayPairSum(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int sum = 0;
        for (int i = 0; i != nums.size(); i += 2)
            sum += nums[i];
        return sum;
    }
};
// 98 ms

[LeetCode] Array Partition I

原文：http://www.cnblogs.com/immjc/p/7137962.html