[LeetCode] Array Partition I

阿新 • • 發佈：2017-07-08

int pla ger log block turn osi you logs

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:

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

給定一個含有2n個整數的數組，將這2n個整數分成n個整數對，對每個整數對中的較小數求和，並使這個和最大。根據題意，要使總和最大，在整個數組中需要每個整數對中較小數的取值盡可能大。先將數組排序，此時把兩兩相鄰的整數作為一個整數對，求每個整數對較小數的和，即求排序後數組奇數次序的數值之和。

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

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