LeetCode——Array Partition I（561）

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].

分析：

实际上是排序然后隔项求和。快排。

static auto _____ = []() {
    std::ios::sync_with_stdio(false);
    cin.tie(NULL);
    return 0;
}();
class Solution {
public:
    int arrayPairSum(vector<int>& nums) {
        QuickSort(nums, 0, nums.size()-1);
        int sum = 0;
        for(int i = 0;i<nums.size();i+=2)
            sum+=nums[i];
        return sum;
    }
    
    void QuickSort(vector<int>& nums, int left, int right)
    {
        if(left < right)
        {
            int index = Pivot(nums, left, right);
            QuickSort(nums, left, index-1);
            QuickSort(nums, index+1, right);
        }
    }
    
    int Pivot(vector<int>& nums, int left, int right)
    {
        int pivot = nums[left];
        while(left < right)
        {
            while(right>left&&nums[right]>=pivot) --right;
            nums[left] = nums[right];
            while(right>left&&nums[left]<=pivot) ++left;
            nums[right] = nums[left];
        }
        nums[left] = pivot;
        return left;
    }
};

后来看到一种更不错的，用来bitmap的思想，免去了排序，看题目的答案就好了，这里就不贴出。