You are given n pairs of numbers. In every pair, the first number is always smaller than the second number.
Now, we define a pair (c, d) can follow another pair (a, b) if and only if b < c. Chain of pairs can be formed in this fashion. 
Given a set of pairs, find the length longest chain which can be formed. You needn't use up all the given pairs. You can select pairs in any order.
Example 1:
Input: [[1,2], [2,3], [3,4]]
Output: 2
Explanation: The longest chain is [1,2] -> [3,4]
Note:
The number of given pairs will be in the range [1, 1000].
class Solution:
     def findLongestChain(self, pairs):
         """
         :type pairs: List[List[int]]
         :rtype: int
         """
         # pairs = sorted(pairs, key=lambda x: x[0]) # 动态规划, 居然也超时了
         # dp = [1] * len(pairs)
         # for i in range(len(pairs)):
         #     for j in range(i):
         #         if pairs[j][1] < pairs[i][0]:
         #             dp[i] = max(dp[i], dp[j]+1)
         # return max(dp)
         
         pairs.sort(key=lambda x: x[1])
         cur = pairs[0][0] - 1
         ans = 0
         for a, b in pairs:
             if cur < a:
                 cur = b
                 ans += 1
         return ans

class Solution:
     def findLongestChain(self, pairs):
         """
         :type pairs: List[List[int]]
         :rtype: int
         """
         # pairs = sorted(pairs, key=lambda x: x[0]) # 动态规划, 居然也超时了
         # dp = [1] * len(pairs)
         # for i in range(len(pairs)):
         #     for j in range(i):
         #         if pairs[j][1] < pairs[i][0]:
         #             dp[i] = max(dp[i], dp[j]+1)
         # return max(dp)
         
         pairs.sort(key=lambda x: x[1])
         cur = pairs[0][0] - 1
         ans = 0
         for a, b in pairs:
             if cur < a:
                 cur = b
                 ans += 1
         return ans
