LeetCode54 Spiral Matrix

Posted 2020-08-09 wangxiaobao的博客

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]. （Medium）

分析：

题目没有什么复杂的算法要应用，就是一行一列输出，处理好细节即可。

比较清晰的写法是搞一个rowBegin, rowEnd, colBegin, colEnd， 并在处理完一行/一列后更新值，并判断是否仍然满足 rowBegin <= rowEnd && colBegin <= colEnd

代码：

 1 class Solution {
 2 public:
 3     vector<int> spiralOrder(vector<vector<int>>& matrix) {
 4         vector<int> result;
 5         if (matrix.size() == 0) {
 6             return result;    
 7         }
 8         int m = matrix.size(), n = matrix[0].size();
 9         int rowBegin = 0, rowEnd = m - 1, colBegin = 0, colEnd = n - 1;
10         while (rowBegin <= rowEnd && colBegin <= colEnd) {
11             for (int i = colBegin; i <= colEnd; ++i ) {
12                 result.push_back(matrix[rowBegin][i]);
13             }
14             rowBegin++;
15             if (rowBegin > rowEnd) {
16                 break;
17             }
18             for (int i = rowBegin; i <= rowEnd; ++i) {
19                 result.push_back(matrix[i][colEnd]);
20             }
21             colEnd--;
22             if (colBegin > colEnd) {
23                 break;
24             }
25             for (int i = colEnd; i >= colBegin; --i) {
26                 result.push_back(matrix[rowEnd][i]);
27             }
28             rowEnd--;
29             if (rowBegin > rowEnd) {
30                 break;
31             }
32             for (int i = rowEnd; i>= rowBegin; --i) {
33                 result.push_back(matrix[i][colBegin]);
34             }
35             colBegin++;
36         }
37         return result;
38     }
39 };