54.Spiral Matrix

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].

 

好久没做题了,博客也好久没更新，今天来一发水题,话说这题有点像poj的滑雪题.

传送门POJ:

题解:http://www.cnblogs.com/dick159/p/5258943.html

 

先说一下这题吧,就是分四个方向,然后从外到里按照顺序遍历,要注意结束条件.(我设置成访问过的值就不访问了)

ps:要判断一下间隔了的2个方向是否发生过移动，如果没有发生过移动则表示遍历已完毕.(这里我就用list的大小是否发生了变化来判断.)

目前没有想到更好的方法。【太笨了(╯﹏╰)】

 

Java Code:

public class Solution {        public List
      
        spiralOrder( int[][] matrix ) {            List
       
         list = new ArrayList
        
         ();            if( matrix == null || matrix.length == 0 )                return list;            int m = matrix[ 0 ].length;            int n = matrix.length;            if( matrix[ 0 ].length == 1 ) {                for( int i = 0; i < n; i++ ) {                    list.add( matrix[ i ][ 0 ] );                }                return list;            }                if( matrix.length == 1 ) {                for( int i = 0; i < m; i++ ) {                    list.add( matrix[ 0 ][ i ] );                }                return list;            }            int l = 0;            int preSize = 0;            int i = -1;            while( matrix[ l ][ l ] != -Integer.MAX_VALUE ) {                preSize = list.size();                for( i = l; i < m - l; i++ ) {                    list.add( matrix[ l ][ i ] );                    matrix[ l ][ i ] = -Integer.MAX_VALUE;                }                if(preSize == list.size()){                    return list;                }                preSize = list.size();                for( i = l + 1; i < n - l; i++ ) {                    list.add( matrix[ i ][ m - l - 1 ] );                    matrix[ i ][ m - l - 1 ] = -Integer.MAX_VALUE;                }                if(preSize == list.size()){                    return list;                }                preSize = list.size();                for( i = m - l - 2; i >= l; i-- ) {                    list.add( matrix[ n - l - 1 ][ i ] );                    matrix[ n - l - 1 ][ i ] = -Integer.MAX_VALUE;                }                if(preSize == list.size()){                    return list;                }                preSize = list.size();                for( i = n - l - 2; i >= l + 1; i-- ) {                    list.add( matrix[ i ][ l ] );                     matrix[ i ][ l ] = -Integer.MAX_VALUE;                }                l++;            }            return list;        }}