2021-03-09:在一個數組中,一個數左邊比它小的數的總和,叫數的小和,全部數的小和累加起來,叫數組小和。求數組小和。
例子: [1,3,4,2,5],1左邊比1小的數:沒有,3左邊比3小的數:1,4左邊比4小的數:一、3,2左邊比2小的數:1,5左邊比5小的數:一、三、四、 2,因此數組的小和爲1+1+3+1+1+3+4+2=16 。java
福哥答案2021-03-09:git
1.歸併排序,從左往右,相等拷右。有代碼。
2.歸併排序模板。有代碼。github
代碼用golang編寫,代碼以下:golang
package main import "fmt" func main() { if true { arr := []int{ 1, 3, 4, 2, 5} ret := smallSum1(arr) fmt.Println("1.從左往右,相等拷右:", ret) } if true { arr := []int{ 1, 3, 4, 2, 5} ret := smallSum2(arr) fmt.Println("2.歸併排序模板:", ret) } } func smallSum1(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 } //求中點 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 { //輔助數組 help := make([]int, R-L+1) i := 0 p1 := L p2 := M + 1 //誰小拷貝誰 ans := 0 for p1 <= M && p2 <= R { if arr[p1] < arr[p2] { ans += (R - p2 + 1) * arr[p1] 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++ } //輔助數組拷貝到原數組 copy(arr[L:R+1], help) return ans } func smallSum2(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 } //求中點 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 { //新增的代碼 ans := 0 windowR := M + 1 for i := M; i >= L; i-- { for windowR >= M+1 && arr[i] < arr[windowR] { windowR-- } ans += (R - windowR) * arr[i] } //輔助數組 help := make([]int, R-L+1) i := 0 p1 := L p2 := M + 1 //誰小拷貝誰 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++ } //輔助數組拷貝到原數組 copy(arr[L:R+1], help) return ans }
執行結果以下:
數組