Leetcode: 315. Count of Smaller Numbers After Self

Url : https://leetcode.com/problems/count-of-smaller-numbers-after-self/
You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i].

Example:

Input: [5,2,6,1]
Output: [2,1,1,0]
Explanation:
To the right of 5 there are 2 smaller elements (2 and 1).
To the right of 2 there is only 1 smaller element (1).
To the right of 6 there is 1 smaller element (1).
To the right of 1 there is 0 smaller element.

解題思路：
該題感覺是逆序對的一個變種，我們再解決逆序對的問題上做一個關係映射，記錄之前元素在什麼位置，然後將逆序對的值增加到相應的位置即可。
代碼：

func countSmaller(nums []int) []int {
	len := len(nums)
	numsHelper := make([]int,len)
	count := make([]int,len)
	indexMap := make([]int,len)
	for i:=0;i<len;i++{
		indexMap[i] = i
	}
	indexMapHelper := make([]int,len)
	mergeSort(nums,numsHelper,0,len-1,count,indexMap,indexMapHelper)
	return count
}

func mergeSort(nums []int,numsHelper []int,start int,end int, count []int,indexMap []int,indexMapHelper []int) {
	if start >= end {
		return
	}
	mid := (start+end)/2
	mergeSort(nums,numsHelper,start,mid,count,indexMap,indexMapHelper)
	mergeSort(nums,numsHelper,mid+1,end,count,indexMap,indexMapHelper)

	i:=start
	j:=mid+1
	tmpCount := 0
	k := start
	for ;i<=mid&&j<=end; {
		if nums[i] > nums[j] {
			numsHelper[k] = nums[j]
			tmpCount++
			indexMapHelper[k] = indexMap[j]
			j++
		} else {
			numsHelper[k] = nums[i]
			count[indexMap[i]] += tmpCount
			indexMapHelper[k] = indexMap[i]
			i++
		}
		k++
	}
	for ;i<=mid;i++ {
		numsHelper[k] = nums[i]
		count[indexMap[i]] += tmpCount
		indexMapHelper[k] = indexMap[i]
		k++
	}
	for ;j<=end;j++ {
		numsHelper[k] = nums[j]
		indexMapHelper[k]= indexMap[j]
		k++
	}

	for m := start;m<=end;m++ {
		nums[m] = numsHelper[m]
		indexMap[m] = indexMapHelper[m]
	}
}