[LeetCode]Spiral Matrix 54

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:http://poj.org/problem?id=1088

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<Integer> spiralOrder( int[][] matrix ) {
            List<Integer> list = new ArrayList<Integer>();
            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;
        }
}