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- algorithms
- Medium (75.70%)
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- Testcase Example: '[[1,2,3],[4,5,6],[7,8,9]]'
给定一个 n × n 的二维矩阵 matrix
表示一个图像。请你将图像顺时针旋转 90 度。
你必须在 原地 旋转图像,这意味着你需要直接修改输入的二维矩阵。请不要 使用另一个矩阵来旋转图像。
示例 1:

输入:matrix = [[1,2,3],[4,5,6],[7,8,9]] 输出:[[7,4,1],[8,5,2],[9,6,3]]
示例 2:

输入:matrix = [[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]]
提示:
n == matrix.length == matrix[i].length
1 <= n <= 20
-1000 <= matrix[i][j] <= 1000
/*
* @lc app=leetcode.cn id=48 lang=javascript
*
* [48] 旋转图像
* 水平翻转+对角线翻转
*/
// @lc code=start
/**
* @param {number[][]} matrix
* @return {void} Do not return anything, modify matrix in-place instead.
*/
var rotate = function (matrix) {
const n = matrix.length
for (let i = 0; i < Math.floor(n / 2); i++) {
for (let j = 0; j < n; j++) {
const tmp = matrix[i][j]
matrix[i][j] = matrix[n - i - 1][j]
matrix[n - i - 1][j] = tmp
}
}
for (let i = 0; i < n; i++) {
for (let j = i + 1; j < n; j++) {
const tmp = matrix[i][j]
matrix[i][j] = matrix[j][i]
matrix[j][i] = tmp
}
}
}
// @lc code=end
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/*
* @lc app=leetcode.cn id=48 lang=javascript
*
* [48] 旋转图像
* 直接找规律,每四个数为一个循环。利用临时变量交换数据
*/
// @lc code=start
/**
* @param {number[][]} matrix
* @return {void} Do not return anything, modify matrix in-place instead.
*/
var rotate = function (matrix) {
const n = matrix.length
// i,j
// j,n-i-1
// n-i-1 n-j-1
// n-j-1,i
for (let i = 0; i < Math.floor(n / 2); i++) {
for (let j = 0; j < Math.floor((n + 1) / 2); j++) {
const 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
}
}
}
// @lc code=end
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