【LeetCode】048. Rotate Image

最新推荐文章于 2025-12-08 10:47:21 发布

转载最新推荐文章于 2025-12-08 10:47:21 发布 · 61 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/Atanisi/p/7492030.html

文章标签：

#数据结构与算法

本文介绍了一种不使用额外空间旋转二维矩阵的方法，并给出了四种不同的实现思路。这些方法通过原地交换矩阵元素来实现顺时针90度旋转，避免了使用额外的数据结构。

题目：

You are given an n x n 2D matrix representing an image.

Rotate the image by 90 degrees (clockwise).

Note:
You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.

Example 1:

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

rotate the input matrix in-place such that it becomes:
[
  [7,4,1],
  [8,5,2],
  [9,6,3]
]

Example 2:

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

rotate the input matrix in-place such that it becomes:
[
  [15,13, 2, 5],
  [14, 3, 4, 1],
  [12, 6, 8, 9],
  [16, 7,10,11]
]

题解：

　　暴力解

Solution 1

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        int n = matrix.size();
        for(int i = 0; i < n / 2; ++i){
            for(int j = i; j < n - 1 - i; ++j){
                int tmp = matrix[i][j];
                matrix[i][j] = matrix[n - 1 - j][i];
                matrix[n - 1 - j][i] = matrix[n - 1 - i][n - 1 - j];
                matrix[n - 1 - i][n - 1 - j] = matrix[j][n - 1 - i];
                matrix[j][n - 1 - i] = tmp;
            }
        }
    }
};

Solution 2

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        int n = matrix.size();
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < n - i; ++j){
                swap(matrix[i][j], matrix[n - 1 - j][n - 1 - i]);
            }
        }
        for(int i = 0; i < n / 2; ++i){
            for(int j = 0; j < n; ++j){
                swap(matrix[i][j], matrix[n - 1 - i][j]);
            }
        }
    }
};

先沿着副对角线（/）翻转，再沿着水平中线翻转。

Solution 3

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        int n = matrix.size();
        for(int i = 0; i < n / 2; ++i){
            for(int j = 0; j < n; ++j){
                swap(matrix[i][j], matrix[n - 1 - i][j]);
            }
        }
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < i; ++j){
                swap(matrix[i][j], matrix[j][i]);
            }
        }
    }
};

先沿着水平中线翻转，再沿着主对角线（\）翻转。

Solution 4

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        int n = matrix.size();
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < i; ++j){
                swap(matrix[i][j], matrix[j][i]);
            }
        }
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < n / 2; ++j){
                swap(matrix[i][j], matrix[i][n - 1 - j]);
            }
        }
    }
};

先对原数组取其转置（即沿着主对角线翻转），然后把每行的数字翻转（沿着竖直中线翻转）。另，Solution 4 也可写为Solution 4.1

Solution 4.1

class Solution {
public:
    void rotate(vector<vector<int> > &matrix) {
        int n = matrix.size();
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                swap(matrix[i][j], matrix[j][i]);
            }
            reverse(matrix[i].begin(), matrix[i].end());
        }
    }
};

转载于:https://www.cnblogs.com/Atanisi/p/7492030.html