LeetCode 646. Maximum Length of Pair Chain (Medium)

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:

1. The number of given pairs will be in the range [1, 1000].

 方法一： stack

思路：把array根据first number和second number从小到大排序

1 维护一个递增栈，栈里存放着的是每一个pair。

2 如果一个pair的first number比栈顶的second number大，那么直接把这个pair加进来。

3 否则如果这个pair的first number比栈顶的second number相等，那么我们直接把这个pair pass掉。

4 如果pair的first number比栈顶的second number大，那么我们弹栈一直到这个栈顶的second number比pair的second number小，然后把这个新的pair加进来。但要注意的是需要判断

是否有弹栈的行为。有弹栈，我们再加pair.

time complexity: O(nlogn)  space complexity: O(n) 

class Solution:
    def findLongestChain(self, pairs: List[List[int]]) -> int:
        if not pairs or len(pairs) == 0: return 0 
        
        pairs.sort(key = lambda x:(x[0], x[1]))
       # print(pairs)
        stack = []
        
        stack.append(pairs[0])
        
        for i in range(1, len(pairs)):
            cur = pairs[i]
            pre = stack[-1]
            
            if cur[0] > pre[1]:
                stack.append(cur)
            elif cur[0] == pre[1]:
                continue 
            else:
                shouldAppend = False
                while stack and stack[-1][1] > cur[1]:
                    stack.pop()
                    shouldAppend = True 
                if shouldAppend:
                    stack.append(cur)
        #print(stack)
        return len(stack)

方法二：greedy

思路：把数组按照second number排序。

1 用两个变量 cur保存当前second number，res保存可以形成的length

2 遍历数组。如果遇到数组的first number比cur大，cur update成这个pair的second number。res+= 1。

time complexity:O(nlogn) space complexity:O(1) 

class Solution:
    def findLongestChain(self, pairs: List[List[int]]) -> int:
        if not pairs or len(pairs) == 0: return 0 
        
        pairs.sort(key = lambda x:(x[1]))
        
        cur = float('-inf')
        res = 0 
        
        for pair in pairs:
            if cur < pair[0]:
                cur = pair[1]
                res += 1 
                
        return res 

方法三：dp

class Solution(object): #Time Limit Exceeded
    def findLongestChain(self, pairs):
        pairs.sort()
        dp = [1] * len(pairs)

        for j in xrange(len(pairs)):
            for i in xrange(j):
                if pairs[i][1] < pairs[j][0]:
                    dp[j] = max(dp[j], dp[i] + 1)

        return max(dp)