题目:
思路:排序--计算交换次数
考虑一下,逆序是说a[i]>a[j],i < j。那么在排序的过程中,会把a[i]和a[j]交换过来,这个交换的过程,每交换一次,就是一个逆序对的“正序”过程。
func InversePairs(nums []int) int { res := mergeSort(nums, 0, len(nums)-1) % 1000000007 return int(res) } func mergeSort(nums []int, start, end int) int64 { if start >= end { return 0 } mid := start + (end - start)/2 cnt := mergeSort(nums, start, mid) + mergeSort(nums, mid + 1, end) tmp := []int{} i, j := start, mid + 1 for i <= mid && j <= end { if nums[i] <= nums[j] { tmp = append(tmp, nums[i]) cnt += int64(j - (mid + 1)) i++ } else { tmp = append(tmp, nums[j]) j++ } } for ; i <= mid; i++ { tmp = append(tmp, nums[i]) cnt += int64(end - (mid + 1) + 1) } for ; j <= end; j++ { tmp = append(tmp, nums[j]) } for i := start; i <= end; i++ { nums[i] = tmp[i - start] } return cnt }
原文:https://www.cnblogs.com/dingxiaoqiang/p/14640524.html