LeetCode 剑指Offer 51.数组中的逆序对

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问题描述

简要思路

使用归并排序的分治思想。
res = merge_sort(nums, l, mid) + merge_sort(nums, mid+1, r) 计算分开后每一部分的逆序对。
res += mid-i+1; 计算两部分之间的逆序对。

ps. 思路来源：y总AcWing归并排序模板题。

代码

class Solution {
    
    
public:
    int temp[50010];

    long long merge_sort(vector<int> &nums, int l, int r){
    
    
        if(l >= r) return 0;
        
        int mid = (l+r)/2;
        long long res = merge_sort(nums, l, mid) + merge_sort(nums, mid+1, r);

        int k = 0, i = l, j = mid+1;
        while(i <= mid && j <= r){
    
    
            if(nums[i] <= nums[j]) temp[k++] = nums[i++];
            else{
    
    
                temp[k++] = nums[j++];
                res += mid-i+1;
            }
        }

        while(i <= mid) temp[k++] = nums[i++];
        while(j <= r) temp[k++] = nums[j++];

        for(i=l, j=0; i <= r; i++,j++) nums[i] = temp[j];

        return res; 
    }

    int reversePairs(vector<int>& nums) {
    
    
        int n = nums.size();
        return merge_sort(nums, 0, n-1);
    }
};