leetcode 561---- ArrayPartition

最新推荐文章于 2022-05-01 11:22:08 发布

原创最新推荐文章于 2022-05-01 11:22:08 发布 · 488 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode 题目

OJ & Leetcode 专栏收录该内容

46 篇文章

订阅专栏

本文探讨了如何将一个包含2n个整数的数组分成n组，每组两个数，使得所有组内的较小数值之和达到最大。通过将数组排序并依次选择每对中的较小值来实现这一目标。

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].

题意，给你一个2n长度数组，分成n个长度为2的数组，使得这n个数组的最小值总和最小

简答题目，把所有的数存入一个排序数组中，每2个数取一个最小值就可以了

int arrayPairSum(int* nums,int numsSize) {
    int i = 0,n=numsSize;
    int sum = 0;
    int store[20001];
    int flag = 0;
    for(i=0;i<20001;i++){
        store[i]=0;
    }
    for(i=0;i<n;i++){
             store[nums[i]+10000]++;
    }
    for(i=0;i<20001;i++){
        if(store[i]>0){
        if(flag == 0){
            sum+= (i-10000);
            store[i]--;
            flag = 1;
            if(store[i]>0){
                i--;
            }
        }else{
            if(store[i]>0){
                store[i]--;
            }
            if(store[i]>0){
                i--;
            }
            flag = 0;
        }
        }else{
            continue;
        }
    }
    return sum;
}