5151
5252<!-- 这里可写通用的实现逻辑 -->
5353
54- 原地旋转,i 的范围是 ` [0, n/2) ` ,j 的范围是 ` [i, n-1-i) ` 。
54+ ** 方法一:原地翻转**
55+ 56+ 根据题目要求,我们实际上需要将 $matrix[ i] [ j ] $ 旋转至 $matrix[ j] [ n - i - 1 ] $。
57+ 58+ 我们可以先对矩阵进行上下翻转,即 $matrix[ i] [ j ] $ 和 $matrix[ n - i - 1] [ j ] $ 进行交换,然后再对矩阵进行主对角线翻转,即 $matrix[ i] [ j ] $ 和 $matrix[ j] [ i ] $ 进行交换。这样就能将 $matrix[ i] [ j ] $ 旋转至 $matrix[ j] [ n - i - 1 ] $ 了。
59+ 60+ 时间复杂度 $O(n^2),ドル其中 $n$ 是矩阵的边长。空间复杂度 $O(1)$。
5561
5662<!-- tabs:start -->
5763
6268``` python
6369class Solution :
6470 def rotate (self matrix : List[List[int ]]) -> None :
65- """
66- Do not return anything, modify matrix in-place instead.
67- """
6871 n = len (matrix)
69- for i in range (n // 2 ):
70- for j in range (i, n - 1 - i):
71- t = matrix[i][j]
72- matrix[i][j] = matrix[n - j - 1 ][i]
73- matrix[n - j - 1 ][i] = matrix[n - i - 1 ][n - j - 1 ]
74- matrix[n - i - 1 ][n - j - 1 ] = matrix[j][n - i - 1 ]
75- matrix[j][n - i - 1 ] = t
72+ for i in range (n >> 1 ):
73+ for j in range (n):
74+ matrix[i][j], matrix[n - i - 1 ][j] = matrix[n - i - 1 ][j], matrix[i][j]
75+ for i in range (n):
76+ for j in range (i):
77+ matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]
7678```
7779
7880### ** Java**
@@ -83,19 +85,63 @@ class Solution:
8385class Solution {
8486 public void rotate (int [][] matrix ) {
8587 int n = matrix. length;
86- for (int i = 0 ; i < n / 2 ; ++ i) {
87- for (int j = i; j < n - 1 - i; ++ j) {
88+ for (int i = 0 ; i < n >> 1 ; ++ i) {
89+ for (int j = 0 ; j < n; ++ j) {
90+ int t = matrix[i][j];
91+ matrix[i][j] = matrix[n - i - 1 ][j];
92+ matrix[n - i - 1 ][j] = t;
93+ }
94+ }
95+ for (int i = 0 ; i < n; ++ i) {
96+ for (int j = 0 ; j < i; ++ j) {
8897 int t = matrix[i][j];
89- matrix[i][j] = matrix[n - j - 1 ][i];
90- matrix[n - j - 1 ][i] = matrix[n - i - 1 ][n - j - 1 ];
91- matrix[n - i - 1 ][n - j - 1 ] = matrix[j][n - i - 1 ];
92- matrix[j][n - i - 1 ] = t;
98+ matrix[i][j] = matrix[j][i];
99+ matrix[j][i] = t;
93100 }
94101 }
95102 }
96103}
97104```
98105
106+ ### ** C++**
107+ 108+ ``` cpp
109+ class Solution {
110+ public:
111+ void rotate(vector<vector<int >>& matrix) {
112+ int n = matrix.size();
113+ for (int i = 0; i < n >> 1; ++i) {
114+ for (int j = 0; j < n; ++j) {
115+ swap(matrix[ i] [ j ] , matrix[ n - i - 1] [ j ] );
116+ }
117+ }
118+ for (int i = 0; i < n; ++i) {
119+ for (int j = 0; j < i; ++j) {
120+ swap(matrix[ i] [ j ] , matrix[ j] [ i ] );
121+ }
122+ }
123+ }
124+ };
125+ ```
126+
127+ ### **Go**
128+
129+ ```go
130+ func rotate(matrix [][]int) {
131+ n := len(matrix)
132+ for i := 0; i < n>>1; i++ {
133+ for j := 0; j < n; j++ {
134+ matrix[i][j], matrix[n-i-1][j] = matrix[n-i-1][j], matrix[i][j]
135+ }
136+ }
137+ for i := 0; i < n; i++ {
138+ for j := 0; j < i; j++ {
139+ matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]
140+ }
141+ }
142+ }
143+ ```
144+ 99145### ** JavaScript**
100146
101147``` js
@@ -104,34 +150,78 @@ class Solution {
104150 * @return {void} Do not return anything, modify matrix in-place instead.
105151 */
106152var rotate = function (matrix ) {
107- const n = matrix .length ;
108- for (let i = 0 ; i < n / 2 ; i++ ) {
109- for (let j = i; j < n - i - 1 ; j++ ) {
110- let t = matrix[i][j];
111- matrix[i][j] = matrix[n - j - 1 ][i];
112- matrix[n - j - 1 ][i] = matrix[n - i - 1 ][n - j - 1 ];
113- matrix[n - i - 1 ][n - j - 1 ] = matrix[j][n - i - 1 ];
114- matrix[j][n - i - 1 ] = t;
153+ matrix .reverse ();
154+ for (let i = 0 ; i < matrix .length ; ++ i) {
155+ for (let j = 0 ; j < i; ++ j) {
156+ [matrix[i][j], matrix[j][i]] = [matrix[j][i], matrix[i][j]];
115157 }
116158 }
117159};
118160```
119161
120- ### ** Go **
162+ ### ** TypeScript **
121163
122- ``` go
123- func rotate (matrix [][]int ) {
124- n := len (matrix)
125- r , c := n/2 , (n+1 )/2
126- for i := 0 ; i < r; i++ {
127- for j := 0 ; j < c; j++ {
128- temp := matrix[i][j]
129- matrix[i][j] = matrix[n-j-1 ][i]
130- matrix[n-j-1 ][i] = matrix[n-i-1 ][n-j-1 ]
131- matrix[n-i-1 ][n-j-1 ] = matrix[j][n-i-1 ]
132- matrix[j][n-i-1 ] = temp
133- }
134- }
164+ ``` ts
165+ /**
166+ Do not return anything, modify matrix in-place instead.
167+ */
168+ function rotate(matrix : number [][]): void {
169+ matrix .reverse ();
170+ for (let i = 0 ; i < matrix .length ; ++ i ) {
171+ for (let j = 0 ; j < i ; ++ j ) {
172+ const = matrix [i ][j ];
173+ matrix [i ][j ] = matrix [j ][i ];
174+ matrix [j ][i ] = t ;
175+ }
176+ }
177+ }
178+ ```
179+ 180+ ### ** C#**
181+ 182+ ``` cs
183+ public class Solution {
184+ public void Rotate (int [][] matrix ) {
185+ int n = matrix .Length ;
186+ for (int i = 0 ; i < n >> 1 ; ++ i ) {
187+ for (int j = 0 ; j < n ; ++ j ) {
188+ int t = matrix [i ][j ];
189+ matrix [i ][j ] = matrix [n - i - 1 ][j ];
190+ matrix [n - i - 1 ][j ] = t ;
191+ }
192+ }
193+ for (int i = 0 ; i < n ; ++ i ) {
194+ for (int j = 0 ; j < i ; ++ j ) {
195+ int t = matrix [i ][j ];
196+ matrix [i ][j ] = matrix [j ][i ];
197+ matrix [j ][i ] = t ;
198+ }
199+ }
200+ }
201+ }
202+ ```
203+ 204+ ### ** Rust**
205+ 206+ ``` rust
207+ impl Solution {
208+ pub fn rotate (matrix : & mut Vec <Vec <i32 >>) {
209+ let n = matrix . len ();
210+ for i in 0 .. n / 2 {
211+ for j in 0 .. n {
212+ let t = matrix [i ][j ];
213+ matrix [i ][j ] = matrix [n - i - 1 ][j ];
214+ matrix [n - i - 1 ][j ] = t ;
215+ }
216+ }
217+ for i in 0 .. n {
218+ for j in 0 .. i {
219+ let t = matrix [i ][j ];
220+ matrix [i ][j ] = matrix [j ][i ];
221+ matrix [j ][i ] = t ;
222+ }
223+ }
224+ }
135225}
136226```
137227
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