2021-03-08：在一个数组中，任何一个前面的数a，和任何一个后面的数b，如果(a,b)是降序的，就称为逆序对。返回逆序对个数。

福哥答案2021-03-08：

1.归并排序，从右往左，相等拷右。有代码。
2.归并排序模板。有代码。

代码用golang编写，代码如下：

package main

import "fmt"

func main() {
    if true {
        arr := []int{3, 1, 7, 0, 2}
        ret := reverPairNumber1(arr)
        fmt.Println("1.从右往左，相等拷右：", ret)
    }
    if true {
        arr := []int{3, 1, 7, 0, 2}
        ret := reverPairNumber2(arr)
        fmt.Println("2.归并排序模板：", ret)
    }
}
func reverPairNumber1(arr []int) int {
    arrLen := len(arr)
    if arrLen <= 1 {
        return 0
    }
    return process1(arr, 0, arrLen-1)
}
func process1(arr []int, L int, R int) int {
    curLen := R - L + 1
    if curLen <= 1 {
        return 0
    }

    //求中点
    M := L + (R-L)>>1
    return process1(arr, L, M) + process1(arr, M+1, R) + merge1(arr, L, M, R)

}
func merge1(arr []int, L int, M int, R int) int {

    //辅助数组
    help := make([]int, R-L+1)
    i := R - L + 1 - 1
    p1 := M
    p2 := R
    res := 0
    //谁小拷谁,相等拷右
    for p1 >= L && p2 >= M+1 {
        if arr[p1] > arr[p2] {
            res += p2 - M
            help[i] = arr[p1]
            p1--
        } else {
            help[i] = arr[p2]
            p2--
        }
        i--
    }

    for p1 >= L {
        help[i] = arr[p1]
        p1--
        i--
    }
    for p2 >= M+1 {
        help[i] = arr[p2]
        p2--
        i--
    }

    //辅助数组拷贝到原数组
    copy(arr[L:R+1], help)
    return res
}

func reverPairNumber2(arr []int) int {
    arrLen := len(arr)
    if arrLen <= 1 {
        return 0
    }
    return process2(arr, 0, arrLen-1)
}
func process2(arr []int, L int, R int) int {
    curLen := R - L + 1
    if curLen <= 1 {
        return 0
    }

    //求中点
    M := L + (R-L)>>1
    return process2(arr, L, M) + process2(arr, M+1, R) + merge2(arr, L, M, R)

}
func merge2(arr []int, L int, M int, R int) int {
    //新增的代码
    ans := 0
    windowR := M + 1
    for i := L; i <= M; i++ {
        for windowR <= R && arr[i] > arr[windowR] {
            windowR++
        }
        ans += windowR - M - 1
    }

    //辅助数组
    help := make([]int, R-L+1)
    i := 0
    p1 := L
    p2 := M + 1
    //谁小拷贝谁
    for p1 <= M && p2 <= R {
        if arr[p1] <= arr[p2] {
            help[i] = arr[p1]
            p1++
        } else {
            help[i] = arr[p2]
            p2++
        }
        i++
    }

    for p1 <= M {
        help[i] = arr[p1]
        p1++
        i++
    }
    for p2 <= R {
        help[i] = arr[p2]
        p2++
        i++
    }

    //辅助数组拷贝到原数组
    copy(arr[L:R+1], help)
    return ans
}