Spiral Matrix

旋转矩阵

题目

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

Example 1:

Input:
[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]
Output: [1,2,3,6,9,8,7,4,5]
Example 2:

Input:
[
[1, 2, 3, 4],
[5, 6, 7, 8],
[9,10,11,12]
]
Output: [1,2,3,4,8,12,11,10,9,5,6,7]

解析重点

1.我们查看矩阵，旋转矩阵的时候会涉及到4个点，上top下bottom左left右right。当我们第一次旋转时，left->right，将这些点旋转完后，top将会受到影响，top+1
类似的，top->bottom，right->left,boottom->top，也会有同样的规律。那么我们什么时候终止呢，当top和bottom，left和right重合时我们就可以终止了。

java代码

public class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        List<Integer> res = new ArrayList<Integer>();
        if(matrix.length == 0 || matrix[0].length == 0) return res;
        
        int top = 0;
        int bottom = matrix.length-1;
        int left = 0;
        int right = matrix[0].length-1;
        
        while(true){
            for(int i = left; i <= right; i++) res.add(matrix[top][i]);
            top++;
            if(left > right || top > bottom) break;
            
            for(int i = top; i <= bottom; i++) res.add(matrix[i][right]);
            right--;
            if(left > right || top > bottom) break;
            
            for(int i = right; i >= left; i--) res.add(matrix[bottom][i]);
            bottom--;
            if(left > right || top > bottom) break;
            
            for(int i = bottom; i >= top; i--) res.add(matrix[i][left]);
            left++;
            if(left > right || top > bottom) break;
        }
        
        return res;
    }
    
}