[leetcode]Spiral Matrix

题目描述如下：

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].

一道简单的模拟题，主要的思路有那么几个：

1、定义一个标志行进方向的变量direction，取值范围为[0,3]，分别代表四个方向；

2、定义两个数组directionX[]，directionY[]，组合使用表示下一个的位置；

3、定义一个标记数组flagArray[]，对经过的位置进行标志；

4、定义两个变量currentX ，currentY代表当前所在的位置；

5、判断调转方向的时机（5个条件，具体见代码），并修改行进方向的变量direction。

具体实现：

public class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
         List<Integer> resList = new ArrayList<Integer>();
        int widtn = matrix.length;
        if(widtn == 0) return resList;
        int len = matrix[0].length;
        int [][]flagArray = new int[widtn][len];
        int direction = 0;  // 初始方向
        int currentX = 0, currentY = 0;
        int []directionY = {1, 0, -1, 0};
        int []directionX = {0, 1, 0, -1};
        int count = len * widtn;
        while(count > 0){
            int value = matrix[currentX][currentY];
            resList.add(value);
            flagArray[currentX][currentY] = 1;
            if(currentX + directionX[direction] < widtn &&
                    currentX + directionX[direction] >= 0 &&
                    currentY + directionY[direction] < len &&
                    currentY + directionY[direction] >= 0 &&
                    flagArray[currentX + directionX[direction]][currentY + directionY[direction]] != 1){
                currentX = currentX + directionX[direction];
                currentY = currentY + directionY[direction];
            }else{
                direction = (++ direction) % 4;
                currentX = currentX + directionX[direction];
                currentY = currentY + directionY[direction];
            }
           count --;
        }
        return resList;
    }
}

题目链接：https://leetcode.com/problems/spiral-matrix/