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48.旋转图像.cpp
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70 lines (67 loc) · 1.47 KB
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/*
* @lc app=leetcode.cn id=48 lang=cpp
*
* [48] 旋转图像
*
* https://leetcode-cn.com/problems/rotate-image/description/
*
* algorithms
* Medium (73.72%)
* Likes: 1140
* Dislikes: 0
* Total Accepted: 262.6K
* Total Submissions: 356.2K
* 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 code=start
class Solution {
public:
void rotate(vector<vector<int>>& matrix) {
// 转置矩阵
for(int i=0;i<matrix.size();++i) {
for(int j=0;j<i;++j) {
swap(matrix[i][j],matrix[j][i]);
}
}
// 左右镜像矩阵
for(int i=0;i<matrix.size();++i) {
for(int j=0;j<matrix.size()/2;++j) {
swap(matrix[i][j],matrix[i][matrix.size()-j-1]);
}
}
}
};
// @lc code=end