Given an unsorted array of integers, find the length of longest increasing subsequence.
Your algorithm should run in O(n2) complexity.
给出一个未排序的整数数组，找出最长增长子序列的长度（算法时间复杂度应该是O(N ^ 2)）
Given [10, 9, 2, 5, 3, 7, 101, 18],
The longest increasing subsequence is [2, 3, 7, 101], therefore the length is 4. Note that there may be more than one LIS combination, it is only necessary for you to return the length.
动态规划解。定义dp[i]：以第i个元素结尾的最长子序列长度（不是整个序列的最长子序列），递推关系式：dp[i] = max{dp[j] > dp[i]} + 1 (0 <= j < i )，并且维护一个最大子序列长度，每当dp[i]更新时同时更新最长子序列长度。
class Solution(object): def lengthOfLIS(self, nums): """ :type nums: List[int] :rtype: int """ if len(nums) == 0: return 0 dp = [1] * len(nums) max_len = 1 # 维护一个最大子序列长度 for index_n, n in enumerate(nums): for i in range(index_n): if n > nums[i] and dp[i] + 1 > dp[index_n]: dp[index_n] = dp[i] + 1 max_len = max(max_len, dp[index_n]) return max_len新闻热点
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