LeetCode 561. Array Partition I

最新推荐文章于 2021-12-12 16:53:33 发布

ZayneHuang

最新推荐文章于 2021-12-12 16:53:33 发布

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本文介绍了一道LeetCode题目Array Partition I的解法，任务是将给定的2n个整数分为n组，每组两个整数，使得所有组中较小数之和尽可能大。通过计数排序和合理配对的方法给出了解答，并附带了C++代码实现。

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题目链接：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:

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].

题目解析：排序后将相邻大小的两个数分在一起。n比较小，用计数排序， $O(n)$ 复杂度，但是注意到数值范围有负数，所以离散化一下映射到正数就可以。

代码如下：16ms Accepted beating 100%

精简的代码能优化运行速度hh虽然只是ms级别上的细微差异～但是要追求完美呀

class Solution {
public:
    int arrayPairSum(vector<int>& nums) {
        int cnt[20001] = {0};
        for (int i = 0; i < nums.size(); i++)
            cnt[nums[i] + 10000]++;
        int ans = 0;
        bool stat = true;
        for (int i = 0; i < 20001; )
        {
            if (cnt[i])
            {
                if (stat)
                    ans += i - 10000;
                stat = !stat;
                cnt[i]--;
            }
            else
                i++;
        }
        return ans;
    }
};

static auto _____ = []() {
    std::ios::sync_with_stdio(false);
    cin.tie(NULL);
    return 0;
}();