/*
* @lc app=leetcode id=48 lang=javascript
*
* [48] Rotate Image
*/
/**
* @param {number[][]} matrix
* @return {void} Do not return anything, modify matrix in-place instead.
*/
var rotate = function (matrix) {
// 时间复杂度O(n^2) 空间复杂度O(1)
// 做法: 先沿着对角线翻转,然后沿着水平线翻转
const n = matrix.length;
function swap(arr, [i, j], [m, n]) {
const temp = arr[i][j];
arr[i][j] = arr[m][n];
arr[m][n] = temp;
}
for (let i = 0; i < n - 1; i++) {
for (let j = 0; j < n - i; j++) {
swap(matrix, [i, j], [n - j - 1, n - i - 1]);
}
}
for (let i = 0; i < n / 2; i++) {
for (let j = 0; j < n; j++) {
swap(matrix, [i, j], [n - i - 1, j]);
}
}
};
Python3 Code:
class Solution:
def rotate(self, matrix: List[List[int]]) -> None:
"""
Do not return anything, modify matrix in-place instead.
先做矩阵转置(即沿着对角线翻转),然后每个列表翻转;
"""
n = len(matrix)
for i in range(n):
for j in range(i, n):
matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]
for m in matrix:
m.reverse()
def rotate2(self, matrix: List[List[int]]) -> None:
"""
Do not return anything, modify matrix in-place instead.
通过内置函数zip,可以简单实现矩阵转置,下面的代码等于先整体翻转,后转置;
不过这种写法的空间复杂度其实是O(n);
"""
matrix[:] = map(list, zip(*matrix[::-1]))
CPP Code:
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 - i - 1; ++j) {
int tmp = matrix[i][j];
matrix[i][j] = matrix[N - j - 1][i];
matrix[N - j - 1][i] = matrix[N - i - 1][N - j - 1];
matrix[N - i - 1][N - j - 1] = matrix[j][N - i - 1];
matrix[j][N - i - 1] = tmp;
}
}
}
};