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面试题 01.07. 旋转矩阵 ​

题目 ​

给你一幅由 N × N 矩阵表示的图像，其中每个像素的大小为 4 字节。请你设计一种算法，将图像旋转 90 度。

不占用额外内存空间能否做到？

示例 1:

给定 matrix = 
[
  [1,2,3],
  [4,5,6],
  [7,8,9]
],

原地旋转输入矩阵，使其变为:
[
  [7,4,1],
  [8,5,2],
  [9,6,3]
]
给定 matrix = 
[
  [1,2,3],
  [4,5,6],
  [7,8,9]
],

原地旋转输入矩阵，使其变为:
[
  [7,4,1],
  [8,5,2],
  [9,6,3]
]

示例 2:

给定 matrix =
[
  [ 5, 1, 9,11],
  [ 2, 4, 8,10],
  [13, 3, 6, 7],
  [15,14,12,16]
], 

原地旋转输入矩阵，使其变为:
[
  [15,13, 2, 5],
  [14, 3, 4, 1],
  [12, 6, 8, 9],
  [16, 7,10,11]
]
给定 matrix =
[
  [ 5, 1, 9,11],
  [ 2, 4, 8,10],
  [13, 3, 6, 7],
  [15,14,12,16]
], 

原地旋转输入矩阵，使其变为:
[
  [15,13, 2, 5],
  [14, 3, 4, 1],
  [12, 6, 8, 9],
  [16, 7,10,11]
]

注意 ：本题与主站 48 题相同：https://leetcode-cn.com/problems/rotate-image/

题解 ​

java
public void rotate(int[][] matrix) {
    int length = matrix.length, temp;
    // [                   [
    //     [1, 2, 3],         [7, 8, 9],
    //     [4, 5, 6],  ==>    [4, 5, 6],
    //     [7, 8, 9]          [1, 2, 3]
    // ]                   ]
    // 上下翻转
    for (int i = 0; i < length / 2; i++) {
        for (int j = 0; j < length; j++) {
            temp = matrix[i][j];
            matrix[i][j] = matrix[length - 1 - i][j];
            matrix[length - 1 - i][j] = temp;
        }
    }

    // [                   [
    //     [7, 8, 9],         [7, 4, 1],
    //     [4, 5, 6],  ==>    [8, 5, 2],
    //     [1, 2, 3]          [9, 6, 3]
    // ]                   ]
    // 右上左下(/)对角线翻转
    for (int i = 0; i < length; i++) {
        for (int j = i + 1; j < length; j++) {
            temp = matrix[i][j];
            matrix[i][j] = matrix[j][i];
            matrix[j][i] = temp;
        }
    }
}
public void rotate(int[][] matrix) {
    int length = matrix.length, temp;
    // [                   [
    //     [1, 2, 3],         [7, 8, 9],
    //     [4, 5, 6],  ==>    [4, 5, 6],
    //     [7, 8, 9]          [1, 2, 3]
    // ]                   ]
    // 上下翻转
    for (int i = 0; i < length / 2; i++) {
        for (int j = 0; j < length; j++) {
            temp = matrix[i][j];
            matrix[i][j] = matrix[length - 1 - i][j];
            matrix[length - 1 - i][j] = temp;
        }
    }

    // [                   [
    //     [7, 8, 9],         [7, 4, 1],
    //     [4, 5, 6],  ==>    [8, 5, 2],
    //     [1, 2, 3]          [9, 6, 3]
    // ]                   ]
    // 右上左下(/)对角线翻转
    for (int i = 0; i < length; i++) {
        for (int j = i + 1; j < length; j++) {
            temp = matrix[i][j];
            matrix[i][j] = matrix[j][i];
            matrix[j][i] = temp;
        }
    }
}