[Leetcode] 498. Diagonal Traverse 解题报告

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:

1. 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;    }};