1424. 对角线遍历 II

题意

题目链接

思路

数组左边有值，才可能右边有值；
然后遍历

代码

class Solution {
public:vector<int> findDiagonalOrder(vector<vector<int>>& nums) {vector<int> ans;vector< pair<int, int> > now;int i = 0;while (i < nums.size() || !now.empty()) {vector<pair<int, int> > tmp;if (i < nums.size()){ans.push_back(nums[i][0]);tmp.push_back(make_pair(i, 0));}i++;for (auto &index:now){int x = index.first;int y = index.second + 1;if (x < nums.size() && y < nums[x].size()){tmp.push_back(make_pair(x, y));ans.push_back(nums[x][y]);}}now = tmp;}return ans;}
};