Leetcode-Spiral Matrix

最新推荐文章于 2021-09-18 15:12:15 发布

转载最新推荐文章于 2021-09-18 15:12:15 发布 · 50 阅读

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原文链接：http://www.cnblogs.com/lishiblog/p/4102804.html

本文介绍了一种矩阵螺旋遍历算法的两种实现方案。第一种使用布尔矩阵跟踪已访问元素，第二种通过更新行列边界简化判断逻辑，提高了效率。

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 }

转载于:https://www.cnblogs.com/lishiblog/p/4102804.html