Leetcode 48. Rotate Image 自制答案
2017-08-02 15:00
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Description
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?
第一版答案 递归
最简单的想法,就是由外到内一层一层的旋转矩阵中的每个元素。1 2 3 4 13 9 5 1 13 9 5 1 5 6 7 8 ---> 14 6 7 2 ---> 14 10 6 2 9 10 11 12 15 10 11 3 15 11 7 3 13 14 15 16 16 12 8 4 16 12 8 4
代码如下:
public void rotate(int[][] matrix) { if(matrix.length==0||matrix.length==1) return; rotate(matrix, 0, matrix.length-1); } public void rotate(int[][] matrix, int start, int end){ if((end - start)<=0) return ; int temp = 0; for(int i=0;i<(end-start);i++){ temp = matrix[end-i][start]; matrix[end-i][start] = matrix[end][end-i]; matrix[end][end-i] = matrix[start+i][end]; matrix[start+i][end] = matrix[start][start+i]; matrix[start][start+i] = temp; } rotate(matrix, start+1, end-1); }
第二版答案 非递归
与上述的答案采用相同的方法,只是不采用递归的方法,减少了运行时间。public void rotate(int[][] matrix) { if(matrix.length==0||matrix.length==1) return; int count = matrix.length; for(int i = 0; i < count / 2; i++){ for(int j = 0; j < (count+1) / 2; j++){ int temp = matrix[count-j-1][i]; matrix[count-j-1][i] = matrix[count-i-1][count-j-1]; matrix[count-i-1][count-j-1] = matrix[j][count-i-1]; matrix[j][count-i-1] = matrix[i][j]; matrix[i][j] = temp; } } }
第三版答案
refer:AC Java in place solution with explanation Easy to understand.如果矩阵顺时针旋转九十度,可以看做是如下两步操作:1、沿着对角线对称翻折。2、左右对折
1 2 3 1 4 7 7 4 1 4 5 6 --> 2 5 8 --> 8 5 2 7 8 9 3 6 9 9 6 3
代码如下:
public void rotate(int[][] matrix) { for(int i = 0; i<matrix.length; i++){ for(int j = i; j<matrix[0].length; j++){ int temp = 0; temp = matrix[i][j]; matrix[i][j] = matrix[j][i]; matrix[j][i] = temp; } } for(int i =0 ; i<matrix.length; i++){ for(int j = 0; j<matrix.length/2; j++){ int temp = 0; temp = matrix[i][j]; matrix[i][j] = matrix[i][matrix.length-1-j]; matrix[i][matrix.length-1-j] = temp; } } }
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