Leetcode-Spiral Matrix

LiBlog發表於2014-11-17
LeetCode

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

For example,
Given the following matrix: 

[
 [ 1, 2, 3 ],
 [ 4, 5, 6 ],
 [ 7, 8, 9 ]
]

You should return [1,2,3,6,9,8,7,4,5].

Solution:

 1 public class Solution {
 2     public List<Integer> spiralOrder(int[][] matrix) {
 3         List<Integer> res = new ArrayList<Integer>();
 4         int xLen = matrix.length;
 5         if (xLen==0) return res;
 6         int yLen = matrix[0].length;
 7         if (yLen==0) return res;
 8 
 9         int[] x = new int[]{0,1,0,-1};
10         int[] y = new int[]{1,0,-1,0};
11         boolean[][] printed = new boolean[xLen][yLen];
12         for (int i=0;i<xLen;i++)
13             Arrays.fill(printed[i],false);
14         int direction = 0;
15         int curX = 0, curY=0;        
16         for (int i=0;i<xLen*yLen;i++){
17             res.add(matrix[curX][curY]);
18             printed[curX][curY] = true;
19             int nextX = curX+x[direction];
20             int nextY = curY+y[direction]; 
21             //Determin the availability of next point.
22             if (nextX>=xLen || nextX<0 || nextY>=yLen || nextY<0 || printed[nextX][nextY]){
23                 direction = (direction+1)%4;
24                 nextX = curX+x[direction];
25                 nextY = curY+y[direction];
26             }
27             curX = nextX;
28             curY = nextY;
29         }
30         return res;        
31     }
32 }

Solution 2:

We actually does not need the matrix to record the visited element, because each time the direction is changed, one row or one col becomes unavailable. We only need to record and update the start and end of the available rows and clos. 

 1 public class Solution {
 2     public List<Integer> spiralOrder(int[][] matrix) {
 3         List<Integer> res = new ArrayList<Integer>();
 4         int xLen = matrix.length;
 5         if (xLen==0) return res;
 6         int yLen = matrix[0].length;
 7         if (yLen==0) return res;
 8 
 9         int[] x = new int[]{0,1,0,-1};
10         int[] y = new int[]{1,0,-1,0};      
11         int direction = 0;
12         int curX = 0, curY=0;        
13         int rowStart = 0, rowEnd = xLen-1, colStart=0, colEnd=yLen-1;
14 
15         for (int i=0;i<xLen*yLen;i++){
16             res.add(matrix[curX][curY]);
17             int nextX = curX+x[direction];
18             int nextY = curY+y[direction]; 
19             //Determin the availability of next point.
20             if (nextX>rowEnd || nextX<rowStart || nextY>colEnd || nextY<colStart){
21                 direction = (direction+1)%4;
22                 nextX = curX+x[direction];
23                 nextY = curY+y[direction];
24                 //Update the availability of rows and cols according the direction change.
25                 if (direction==1) rowStart++;
26                 if (direction==2) colEnd--;
27                 if (direction==3) rowEnd--;
28                 if (direction==0) colStart++;
29             }
30             curX = nextX;
31             curY = nextY;
32         }
33         return res;        
34     }
35 }

 

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