<p>Given a 2D integer array <code>nums</code>, return <em>all elements of </em><code>nums</code><em> in diagonal order as shown in the below images</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/04/08/sample_1_1784.png" style="width: 158px; height: 143px;" />
<pre>
<strong>Input:</strong> nums = [[1,2,3],[4,5,6],[7,8,9]]
<strong>Output:</strong> [1,4,2,7,5,3,8,6,9]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/04/08/sample_2_1784.png" style="width: 230px; height: 177px;" />
<pre>
<strong>Input:</strong> nums = [[1,2,3,4,5],[6,7],[8],[9,10,11],[12,13,14,15,16]]
<strong>Output:</strong> [1,6,2,8,7,3,9,4,12,10,5,13,11,14,15,16]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i].length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= sum(nums[i].length) &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i][j] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    vector<int> findDiagonalOrder(vector<vector<int>>& nums) {
        const int n = nums.size();
        list<pair<int, int>> l;
        for (int i = n-1; i >= 0; i--) {
            l.push_back({i, 0});
        }
        vector<int> res;
        int pl = 0;
        auto s = l.end(); s--;
        bool reached = false;
        for (;;) {
            int pl = res.size();
            auto ns = s;
            if (!reached && ns != l.begin()) --ns;

            for (auto li = (reached ? l.begin() : s); li != l.end();) {
                int i = li->first;
                int j = li->second;
                if (j < nums[i].size()) {
                    res.push_back(nums[i][j]);
                    li->second++;
                    li++;
                    // cout << "adv ";
                } else {
                    li = l.erase(li);
                    // cout << "removed ";
                }
                // cout << i << ' ' << c[i] << endl;
            }

            s = ns;
            reached |= (s == l.begin());
/*
            for (auto tp = l.begin(); tp != l.end(); tp++) {
                cout << *tp << " ";
            }
            cout << endl;
*/
            if (pl == res.size()) break;
        }

        return res;
    }
};
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