实现二维数组顺时针旋转的功能

Posted 2021-01-02 hsnblog

用GO实现二维数组的顺时针旋转，总体思想为，确定好正方形二维数组的边界，从边开始旋转，转完整个边界之后把二维数组向内缩小一个边界，找到边界，继续旋转(交换)....

例如：

{ 1,  2,  3,  4}

{ 5,  6,  7,  8}

{ 9,10,11,12}

{13,14,15,16}

旋转后:

{13,  9,  5,  1}

{14,10,  6,  2}

{15,11,  7,  3}

{16,12,  8,  4}

 

 1 package algorithm
 2 
 3 func rotateSquare(arr *[][]int) {
 4     topLeftRow := 0
 5     topLeftCol := 0
 6     lowRightRow := len(*arr) - 1
 7     lowRightCol := len((*arr)[0]) - 1
 8     for ; topLeftRow <= lowRightRow && topLeftCol <= lowRightCol; {
 9         rotateEdge(arr, topLeftRow, topLeftCol, lowRightRow, lowRightCol)
10         topLeftRow++
11         topLeftCol++
12         lowRightRow--
13         lowRightCol--
14     }
15 }
16 
17 func rotateEdge(arr *[][]int, topLeftRow, topLeftCol, lowRightRow, lowRightCol int) {
18     times := lowRightCol - topLeftCol
19     for i := 0; i < times; i++ {
20         origin := (*arr)[topLeftRow][topLeftCol+i]
21         (*arr)[topLeftRow][topLeftCol+i] = (*arr)[lowRightRow-i][topLeftCol]
22         (*arr)[lowRightRow-i][topLeftCol] = (*arr)[lowRightRow][lowRightCol-i]
23         (*arr)[lowRightRow][lowRightCol-i] = (*arr)[topLeftRow+i][lowRightCol]
24         (*arr)[topLeftRow+i][lowRightCol] = origin
25     }
26 }

测试代码为

 1 package algorithm
 2 
 3 import (
 4     "testing"
 5     "fmt"
 6 )
 7 
 8 func Test_rotateSquare(t *testing.T) {
 9     arr := [][]int{
10         {1, 2, 3, 4},
11         {5, 6, 7, 8},
12         {9, 10, 11, 12},
13         {13, 14, 15, 16},
14     }
15     rotateSquare(&arr)
16     fmt.Println(arr)
17 }