问题描述：

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].

基本思路：

思路没什么特殊的，只是最普通的想法。但是值得注意的是处理旋转的细节。

代码：

vector<int> spiralOrder(vector<vector<int> > &matrix) {  //C++
        enum direct{toRight, toDown, toLeft, toUp}; 
        
        //init record
        vector<vector<int> > record;
         vector<int> result;
        int m = matrix.size();
        if(m == 0)
            return result;
            
        int n = matrix[0].size();
        for(int i = 0; i < m; i++){
            vector<int> tmp(n,0);
            record.push_back(tmp);
        }
        
        int allnum = m*n;
        
        
        //walk
        direct x = toRight;
        int i =0 ,j =0;
        result.push_back(matrix[0][0]);
        int count = 1;
        record[0][0] = 1;
        while(count < allnum){
            switch(x){
                case toRight:
                   while(j+1 < n && record[i][j+1] == 0 ){
                        result.push_back(matrix[i][j+1]);
                        count++;
                        record[i][j+1] = 1;
                        j++;
                   }
                   x = toDown;
                   break;
                case toDown:
                    while(i+1 < m &&record[i+1][j] == 0){
                        result.push_back(matrix[i+1][j]);
                        count++;
                        record[i+1][j] = 1;
                        i++;
                    }
                    x = toLeft;
                    break;
                case toLeft:
                    while(j-1>=0 && record[i][j-1]==0){
                        result.push_back(matrix[i][j-1]);
                        count++;
                        record[i][j-1] = 1;
                        j--;
                    }
                    x = toUp;
                    break;
                case toUp:
                    while(i-1 >= 0 && record[i-1][j]==0){
                        result.push_back(matrix[i-1][j]);
                        count++;
                        record[i-1][j] = 1;
                        i--;
                    }
                    x = toRight;
                    break;
                default:
                    break;
            }
        }
        return result;
    }

a better implement in java

public class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {

        List<Integer> res = new ArrayList<Integer>();

        if (matrix.length == 0) {
            return res;
        }

        int rowBegin = 0;
        int rowEnd = matrix.length-1;
        int colBegin = 0;
        int colEnd = matrix[0].length - 1;

        while (rowBegin <= rowEnd && colBegin <= colEnd) {
            // Traverse Right
            for (int j = colBegin; j <= colEnd; j ++) {
                res.add(matrix[rowBegin][j]);
            }
            rowBegin++;

            // Traverse Down
            for (int j = rowBegin; j <= rowEnd; j ++) {
                res.add(matrix[j][colEnd]);
            }
            colEnd--;

            if (rowBegin <= rowEnd) {
                // Traverse Left
                for (int j = colEnd; j >= colBegin; j --) {
                    res.add(matrix[rowEnd][j]);
                }
            }
            rowEnd--;

            if (colBegin <= colEnd) {
                // Traver Up
                for (int j = rowEnd; j >= rowBegin; j --) {
                    res.add(matrix[j][colBegin]);
                }
            }
            colBegin ++;
        }

        return res;
    }
}

[leetcode]Spiral Matrix