Leetcode 300. Longest Increasing Subsequence 动态规划经典题

300. Longest Increasing Subsequence（Medium)

Given an unsorted array of integers, find the length of longest increasing subsequence.
给定一个无序的整数数组，找到其中最长上升子序列的长度。
Example:

Input: [10,9,2,5,3,7,101,18]
Output: 4
Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.
Note:

There may be more than one LIS combination, it is only necessary for you to return the length.Your algorithm should run in O(n2) complexity.
Follow up: Could you improve it to O(n log n) time complexity?

方法一：
动态规划，dp[] 代表该位的最长上升子序列最大长度。遍历一遍前面比自己小的数比较就行。

    def lengthOfLIS(self, nums):
        if not nums:
            return 0
        dp = [1] * len(nums)
        for i in range(1, len(nums)):
            for j in range(0, i):
                if nums[i] > nums[j]:
                    dp[i] = max(dp[i], dp[j] + 1)
        return max(dp)

方法二：
二分搜索法，维护一个最大增长的子序列List res，遇到小的数就替换，需要注意的是只是替换并没有改变位数，只有后面的数大于最后一位数res才会增加。

    def lengthOfLIS(self, nums):
        if not nums:
            return 0
        res = [nums[0]]
        for i in range(1, len(nums)):
            if nums[i] > res[-1]:
                res.append(nums[i])
            else:  #binarysearch 只是替换，并没有增加数
                l, r, mid = 0, len(res) - 1, 0
                while l <= r:
                    mid = (l + r) // 2
                    if res[mid] < nums[i]:
                        l = mid + 1
                    else:
                        r = mid - 1
                res[l] = nums[i]
                print(res)
        return len(res)