1 题目描述
给定一个 n x n 的二维矩阵表征一幅图像,求顺时针旋转90度后的图像。
注:须使用现有内存空间原地旋转该图像,勿分配额外的二维矩阵空间来做旋转。
例子1:
输入矩阵:
[
[1,2,3],
[4,5,6],
[7,8,9]
]
输出:
[
[7,4,1],
[8,5,2],
[9,6,3]
]
例子2:
输入矩阵:
[
[ 5, 1, 9,11],
[ 2, 4, 8,10],
[13, 3, 6, 7],
[15,14,12,16]
],
输出:
[
[15,13, 2, 5],
[14, 3, 4, 1],
[12, 6, 8, 9],
[16, 7,10,11]
]
题目出处:
https://leetcode.com/problems/rotate-image/
2 解决思路
1)如图所示,原始图如下图所示,要不分配额外空间,以最小的空间来作旋转,我们设定每次仅移动一位。
2)一圈元素顺时针旋转90度的移动过程如下图左图所示,仅申请一个元素的存储单元tmp,作顺时针旋转时,我们每次仅移动一位。步骤如下。
2.1)左上角元素移入tmp;
2.1)最左边一列元素均向上移动一位;
2.2)最底部一行元素均向左移动一位;
2.3)最右边一列元素均向下移动一位;
2.4)最顶部一行元素均向右移动一位;
2.5)tmp元素置入顶部行的最左边第二列;
2.6)循环移动n-1次,直至下图右图所示,整个最外围一圈的元素即顺时针90度旋转完成。
3)最外围边界递进一层,开始次外围一圈元素的旋转,移动方式同2-3相同。
4)直至最里边一圈的元素均旋转完成,即得到整个图像的旋转。
3 golang实现代码
https://github.com/leileiluoluo/leetcode/blob/master/48_Rotate_Image/test.go
func rotate(matrix [][]int) {
for level := 0; level < len(matrix)/2; level++ {
topBoundary := level
leftBoundary := level
bottomBoundary := len(matrix) - 1 - level
rightBoundary := len(matrix) - 1 - level
for times := 0; times < bottomBoundary-topBoundary; times++ {
// left
mostLeftTop := matrix[topBoundary][leftBoundary]
for row := topBoundary; row < bottomBoundary; row++ {
matrix[row][leftBoundary] = matrix[row+1][leftBoundary]
}
// bottom
for col := leftBoundary; col < rightBoundary; col++ {
matrix[bottomBoundary][col] = matrix[bottomBoundary][col+1]
}
// right
for row := bottomBoundary; row > topBoundary; row-- {
matrix[row][rightBoundary] = matrix[row-1][rightBoundary]
}
// top
for col := rightBoundary; col > leftBoundary+1; col-- {
matrix[topBoundary][col] = matrix[topBoundary][col-1]
}
matrix[topBoundary][leftBoundary+1] = mostLeftTop
}
}
}
4 基准测试
package main
import "testing"
func BenchmarkRotate(b *testing.B) {
matrix := [][]int{
{5, 1, 9, 11},
{2, 4, 8, 10},
{13, 3, 6, 7},
{15, 14, 12, 16},
}
for i := 0; i < b.N; i++ {
rotate(matrix)
}
}
goos: darwin
goarch: amd64
pkg: github.com/leileiluoluo/test
BenchmarkRotate-4 20000000 65.2 ns/op 0 B/op 0 allocs/op
PASS
ok github.com/leileiluoluo/test 1.383s