[leetcode] 1027. Longest Arithmetic Subsequence
原创
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Description
Given an array A of integers, return the length of the longest arithmetic subsequence in A.
Recall that a subsequence of A is a list A[i_1], A[i_2], …, A[i_k] with 0 <= i_1 < i_2 < … < i_k <= A.length - 1, and that a sequence B is arithmetic if B[i+1] - B[i] are all the same value (for 0 <= i < B.length - 1).
Example 1:
Input: A = [3,6,9,12]
Output: 4
Explanation:
The whole array is an arithmetic sequence with steps of length = 3.
Example 2:
Input: A = [9,4,7,2,10]
Output: 3
Explanation:
The longest arithmetic subsequence is [4,7,10].
Example 3:
Input: A = [20,1,15,3,10,5,8]
Output: 4
Explanation:
The longest arithmetic subsequence is [20,15,10,5].
Constraints:
- 2 <= A.length <= 1000
- 0 <= A[i] <= 500
分析
题目的意思是:找出最长算术子序列,即子序列两数之间的差为定长。这道题的解法也很巧妙,首先是动态规划,但是dp数组里面存放的是字典,字典的键为差值,值为差值的统计量,这一点是要明确的。
接下来就是计算求出最大长度,对于当前的A[i],计算前面的数A[j]与A[i]的差值,然后更新这个dp数组,维护最大长度res就新行了。递推公式为:
dp[i][diff]=max(2,dp[j][diff]+1)
根据题目的意思,等差数列最小长度为2,所以就从2开始了哈哈。
代码
class Solution:
def longestArithSeqLength(self, A: List[int]) -> int:
n=len(A)
d=[defaultdict(int) for i in range(n)]
res=2
for i in range(n):
for j in range(i):
diff=A[i]-A[j]
d[i][diff]=max(2,d[j][diff]+1)
res=max(res,d[i][diff])
return res
参考文献
[LeetCode] Python, DP O(n^2) Solution