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# Solution 1 DP

* t-complexity: $ O(n^2) $
* s-complexity: $ O(n) $

根据数对的第一个数排序所有的数对，dp\[i] 存储以 pairs\[i] 结尾的最长链的长度。当 i < j 且 pairs\[i]\[1] < pairs\[j]\[0] 时，扩展数对链，更新 dp\[j] = max(dp\[j], dp\[i] + 1)。