[Leetcode] 498. Diagonal Traverse 解题报告

本文链接：https://blog.youkuaiyun.com/magicbean2/article/details/78726918

这篇博客介绍了LeetCode第498题的解题报告，题目要求按对角线顺序返回矩阵的所有元素。作者强调了处理边界情况的重要性，并分享了具有O(m*n)时间复杂度和O(1)空间复杂度的解决方案代码。

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题目：

Given a matrix of M x N elements (M rows, N columns), return all elements of the matrix in diagonal order as shown in the below image.

Example:

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

Note:

The total number of elements of the given matrix will not exceed 10,000.

思路：

题目的难度不大，但是需要注意各种边界情况。读者可以重点研究一下for循环里面的四个if语句。算法的时间复杂度是O(m*n)，空间复杂度是O(1)。

代码：

class Solution {
public:
    vector<int> findDiagonalOrder(vector<vector<int>>& matrix) {
        if (matrix.size() == 0 || matrix[0].size() == 0) {
            return {};
        }
        int row_num = matrix.size(), col_num = matrix[0].size();
        vector<int> ret;
        int row = 0, col = 0, d = 0;
        vector<vector<int>> dirs = {{-1, 1}, {1, -1}};
        for (int i = 0; i < row_num * col_num; ++i) {
            ret.push_back(matrix[row][col]);
            row += dirs[d][0], col += dirs[d][1];
            if (row >= row_num) {       // from down to up
                row = row_num - 1, col += 2, d = 1 - d;
            }
            if (col >= col_num) {       // from up to down
                col = col_num - 1, row += 2, d = 1 - d;
            }
            if (row < 0) {              // from up to down
                row = 0, d = 1 - d;
            }
            if (col < 0) {              // from down to up
                col = 0, d = 1 - d;
            }
        }
        return ret;
    }
};