Lintcode: Matrix Zigzag Traversal
Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in ZigZag-order. Have you met this question in a real interview? Yes Example Given a matrix: [ [1, 2, 3, 4], [5, 6, 7, 8], [9,10, 11, 12] ] return [1, 2, 5, 9, 6, 3, 4, 7, 10, 11, 8, 12] Tags Expand
先斜上走到顶,再斜下走到底,直到计数器满, 写的时候老是fail,才发现14行for循环i不需要++,循环里面自己加了
注意corner cases, 以斜上为例
如果是
1,2
5,6
9,10
中6的这种情况,下一个点是10,则x = x+2, y=y-1
如果是1这种情况, 下一个点是2,则只需x = x+1
1 public class Solution { 2 /** 3 * @param matrix: a matrix of integers 4 * @return: an array of integers 5 */ 6 public int[] printZMatrix(int[][] matrix) { 7 // write your code here 8 if (matrix==null || matrix.length==0 || matrix[0].length==0) return null; 9 int m = matrix.length; 10 int n = matrix[0].length; 11 int count = m*n; 12 int[] res = new int[count]; 13 int x=0, y=0; 14 for (int i=0; i<count;) { 15 while (x>=0 && y<n) { 16 res[i++] = matrix[x--][y++]; 17 } 18 if (i == count) break; 19 if (x<0 && y<n) { 20 x++; 21 } 22 else { 23 x = x+2; 24 y = y-1; 25 } 26 while (x<m && y>=0) { 27 res[i++] = matrix[x++][y--]; 28 } 29 if (i == count) break; 30 if (x<m && y<0) { 31 y++; 32 } 33 else { 34 x = x-1; 35 y = y+2; 36 } 37 } 38 return res; 39 } 40 }