Maximum Length of Pair Chain
You are givenn
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 ifb < 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].
分析
因为可以任意顺序,但是每个Pair first<second,先排序
dp表示到当前i 最长的chain, 2层循环,为了及时返回,第二层倒序。
class Solution:
def findLongestChain(self, pairs: List[List[int]]) -> int:
n = len(pairs)
f = [1]*(n+1)
res = 0
# for i in range(1,n+1):
# f[i] = 1
pairs.sort()
for i in range(1,n+1):
for j in range(i-1,-1,-1): #倒序是为了及时break 防止TLE
ca,cb = pairs[i-1]
pa,pb = pairs[j-1]
if ca > pb:
f[i] = max(f[i],f[j] + 1)
break
res = max(res,f[i])
return res
按照end排序,然后一个个数顺移计算res,同时更新end
不按start,因为可能第一个数范围就涵盖了后面所有数,更新的end也大,没有任何start能超过它。导致结果是1,具体解释:
https://leetcode.com/problems/maximum-length-of-pair-chain/discuss/105607/4-Liner-Python-Greedy
class Solution:
def findLongestChain(self, pairs: List[List[int]]) -> int:
pairs.sort(key = lambda x: (x[1],x[0]))#记得要加括号!!!!
pre = float('-inf')
res = 0
for a, b in pairs:
if pre < a:
pre,res = b,res+1
return res
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