【LeetCode】048. Rotate Image
阿新 • • 發佈:2017-09-07
swap cat int directly inpu 對角線 leet bsp log
題目:
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] ]
題解:
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]); } } } };
先沿著副對角線(/)翻轉,再沿著水平中線翻轉。
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]); } } } };
先沿著水平中線翻轉,再沿著主對角線(\)翻轉。
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]); } } } };
先對原數組取其轉置(即沿著主對角線翻轉),然後把每行的數字翻轉(沿著豎直中線翻轉)。另,Solution 4 也可寫為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()); } } };
【LeetCode】048. Rotate Image