561. Array Partition I

标签（空格分隔）： leetcode

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1. 原题

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.

2. 分析

• 2n个元素分为n组, 要使最小值的和最大，考虑到排序后，每相邻的2个元素之间总是取较小者
• 要使和最大，必须要尽量使组间的差较小，这样在取较小者的时候不会浪费大数，所以只要排序，取下标为偶数的就好

3. 代码

class Solution {
public:
    int arrayPairSum(vector<int>& nums) {
        vector<int>::iterator start = nums.begin();
        vector<int>::iterator end = nums.end();

        sort(start, end);
        int i = 0;
        int sum = 0;

        while (i < nums.size()){
            sum += nums[i];
            i+=2;
        }
        return sum;
    }
};