54. Spiral Matrix

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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].

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比较简单，旋转着往里面填数就行，不过这里的方法略繁琐

public class Solution {    public List<Integer> spiralOrder(int[][] matrix) {        int forward = 0;        int i =0,j=0;        List<Integer> ret =new ArrayList<>();        int targetI = 0, targetJ = 0;        if(matrix.length==0)return ret;        int m = matrix.length, n = matrix[0].length;        if(m*n==0)return ret;        if(m*n==1){            ret.add(matrix[0][0]);            return ret;        }        if(m<=n){            targetI=m/2;            if(m%2==1){                targetJ=n-1-targetI;            }else{                targetJ=m-1-targetI;            }        }else{            targetJ=(n-1)/2;            if(n%2==1){                targetI=m-n+targetJ;            }else{                targetI=targetJ+1;            }        }        // outer:        while(true){            switch(forward){                case 0:                    if(j==n-1-i){                        ret.add(matrix[i][j]);                        forward++;                        i++;                    }else{                        ret.add(matrix[i][j]);                        j++;                    }break;                case 1:                    if(i==m-n+j){                        ret.add(matrix[i][j]);                        forward++;                        j--;                    }else{                        ret.add(matrix[i][j]);                        i++;                    }break;                case 2:                    if(j==m-1-i){                        ret.add(matrix[i][j]);                        forward++;                        i--;                    }else{                        ret.add(matrix[i][j]);                        j--;                    }break;                case 3:                    if(i==j+1){                        ret.add(matrix[i][j]);                        forward++;                        j++;                    }else{                        ret.add(matrix[i][j]);                        i--;                    }break;            }            forward%=4;            if(i==targetI&&j==targetJ)break;        }        ret.add(matrix[i][j]);        return ret;    }}

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