剑指 Offer 51. 数组中的逆序对

 

思路#

方法：归并排序求逆序对#

题解思路来自：力扣官方题解 - 数组中的逆序对

 

复杂度分析#

 

class Solution {
public:
    int reversePairs(vector<int>& nums) {
        vector<int> tmp(nums.size());
        return mergeSort(nums, tmp, 0, nums.size()-1);
    }

    int mergeSort(vector<int> &nums, vector<int>& tmp, int L, int R) {
        if(L >= R) {
            return 0;
        }

        int reversePairsNum = 0;
        int mid = (L + R) / 2;
        reversePairsNum = mergeSort(nums, tmp, L, mid) + mergeSort(nums, tmp, mid+1, R);

        int k = 0;
        int i = L, j = mid+1;
        while(i <= mid && j <= R) {
            if(nums[i] <= nums[j]) {
                tmp[k++] = nums[i++];
                reversePairsNum += j-1-mid; //第i个元素和区间[mid+1, j-1]中的每个元素组成一个逆序对
            } else {
                tmp[k++] = nums[j++];
            }
        }

        while(i <= mid) {
            tmp[k++] = nums[i++];
            reversePairsNum += j-1-mid; //注意：执行此循环的时候说明j已经等于R+1了
        }

        while(j <= R) {
            tmp[k++] = nums[j++];   //注意：执行此循环的时候说明i已经等于mid+1了
        }

        copy(tmp.begin(), tmp.begin()+k, nums.begin()+L);

        return reversePairsNum;
    }
};