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You are given an n x n 2D matrix representing an image.
Rotate the image by 90 degrees (clockwise).
Follow up:
Could you do this in-place?
思路一：
b[i][j] = a[n-1-j][i], 构造b，然后将b赋值给a，空间复杂度O（n*n）
class Solution { public: void rotate(vector<vector<int> > &matrix) { size_t n = matrix.size(); vector<vector<int> > re; vector<int> b; b.resize(n); for(int i = 0; i < n; i ++) re.push_back(b); for(int i = 0; i < n; i ++) for(int j = 0; j < n; j ++) { re[i][j] = matrix[n-j-1][i]; } for(int i = 0; i < n; i ++) for(int j = 0; j < n; j ++) matrix[i][j] = re[i][j]; } };
思路二：旋转90°= 第一步， 沿着副对角线翻转+ 第二步，沿着水平中轴线翻转，空间复杂度O（1）
class Solution { public: #if 0 void rotate(vector<vector<int> > &matrix) { size_t n = matrix.size(); vector<vector<int> > re; vector<int> b; b.resize(n); for(int i = 0; i < n; i ++) re.push_back(b); for(int i = 0; i < n; i ++) for(int j = 0; j < n; j ++) { re[i][j] = matrix[n-j-1][i]; } for(int i = 0; i < n; i ++) for(int j = 0; j < n; j ++) matrix[i][j] = re[i][j]; } #endif public: void rotateFromDiagonal (vector<vector<int> > &matrix) { size_t n = matrix.size(); int tmp; for(int i = 0; i < n; i ++) for(int j = 0; j < n-i; j ++) { tmp = matrix[i][j]; matrix[i][j] = matrix[n-j-1][n-i-1]; matrix[n-j-1][n-i-1] = tmp; } } void rotateFromHorizon(vector<vector<int> > &matrix) { size_t n = matrix.size(); int tmp; for(int i = 0; i < n/2; i ++) for(int j = 0; j < n; j ++) { tmp = matrix[i][j]; matrix[i][j] = matrix[n-1-i][j]; matrix[n-1-i][j] = tmp; } } void rotate(vector<vector<int> > &matrix) { rotateFromDiagonal(matrix); rotateFromHorizon(matrix); } };
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原文地址：http://www.cnblogs.com/diegodu/p/4310811.html
