LeetCode 54. Spiral Matrix

原题链接在这里：https://leetcode.com/problems/spiral-matrix/

题目：

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].

题解：

转圈取值. 矩阵size 是 m*n.

r1 = 0, r2 = m - 1, c1 = 0, c2 = n - 1 as boundary.

while loop condition r1 <= r2, and c1 <= c2. As this fullfills, we need to fill the value to res.

When there is only one row or one column left, we only need first 2 for loops. To check this, use r1 < r2 && c1 < c2. 

As when r1 == r2 || c1 == c2, there is only one row or one column left.

Note, the last for loop stop condition is r > r1, not r >= r1 since r == r1 has been covered by the first loop.

Time Complexity: O(m*n).

Space: O(1), regardless res.

AC Java:

class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        List<Integer> res = new ArrayList<>();
        if(matrix == null || matrix.length == 0 || matrix[0].length == 0){
            return res;
        }
        
        int m = matrix.length;
        int n = matrix[0].length;
        int r1 = 0, c1= 0;
        int r2 = m - 1, c2 = n - 1;
        while(r1 <= r2 && c1 <= c2){
            for(int c = c1; c <= c2; c++){
                res.add(matrix[r1][c]);
            }
            
            for(int r = r1 + 1; r <= r2; r++){
                res.add(matrix[r][c2]);
            }
            
            if(r1 < r2 && c1 < c2){
                for(int c = c2 - 1; c >= c1; c--){
                    res.add(matrix[r2][c]);
                }
                
                for(int r = r2 -1; r > r1; r--){
                    res.add(matrix[r][c1]);
                }
            }
            
            r1++;
            r2--;
            c1++;
            c2--;
        }
        
        return res;
    }
}

跟上Spiral Matrix II, Spiral Matrix III, Spiral Matrix IV.

类似Diagonal Traverse.