由递归排序引申出的小和问题和逆序数对问题

递归排序的思想是将带牌列数组不断的拆成俩部分，再将有序的左右的俩部分合并。

小和问题：在一个数组中，每一个数左边比当前小的数累加起来，叫做这个数组的小和。求一个数组的小和。

逆序对问题，在一个数组中，打印出数组所有左边数大于右边数的组合。

下面是归并排序的代码（go语言）

// 写一个归并排序
func mergeSort(arr *[]int, l, r int) {
	// 这部分是拆
	if l == r {
		return
	}
	mid := l + ((r - l) >> 1)
	mergeSort(arr, l, mid)
	mergeSort(arr, mid+1, r)
	merge(arr, l, mid, r)
}

func merge(arr *[]int, l, mid, r int) {

	p, q := l, mid+1
	help := make([]int, r-l+1)
	i := 0
	for p <= mid && q <= r {
		if (*arr)[p] <= (*arr)[q] {
			help[i] = (*arr)[p]
			p = p + 1
			i = i + 1
		} else {
			help[i] = (*arr)[q]
			q = q + 1
			i = i + 1
		}
	}
	for p <= mid {
		help[i] = (*arr)[p]
		p = p + 1
		i = i + 1
	}
	for q <= r {
		help[i] = (*arr)[q]
		q = q + 1
		i = i + 1
	}
	for i := 0; i < len(help); i++ {
		(*arr)[l+i] = help[i]
	}

}

下面是解决小和问题和逆序数对问题的代码：

func smallSum(arr []int, l, r int) int {
	if len(arr) == 0 || l == r {
		return 0
	}
	mid := l + ((r - l) >> 1)
	return smallSum(arr, l, mid) + smallSum(arr, mid+1, r) + merge2(arr, l, mid, r)
}

func merge2(arr []int, l, mid, r int) int {

	p, q := l, mid+1
	i := 0
	result := 0
	help := make([]int, r-l+1)
	for p <= mid && q <= r {
		if arr[p] < arr[q] {
			result = result + arr[p]*(r-q+1)
			help[i] = arr[p]
			p = p + 1
			i = i + 1
		} else {
			help[i] = arr[q]
			q = q + 1
			i = i + 1
		}
	}
	for p <= mid {
		help[i] = arr[p]
		p = p + 1
		i = i + 1
	}
	for q <= r {
		help[i] = arr[q]
		i = i + 1
		q = q + 1
	}
	p, q = l, mid+1
	for p <= mid {
		for arr[p] > arr[q] {
			fmt.Println(arr[p], arr[q])
			if q < r {
				q = q + 1
			} else {
				break
			}
		}
		p = p + 1
		q = mid + 1
	}
	for i := 0; i < len(help); i++ {
		arr[l+i] = help[i]
	}

	return result
}