leetcode Longest Increasing Subsequence

leetcode Longest Increasing Subsequence

Given an unsorted array of integers, find the length of longest increasing subsequence.

For example, 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.

Your algorithm should run in O(n2) complexity.

Follow up: Could you improve it to O(n log n) time complexity?

题目地址：leetcode Longest Increasing Subsequence

题意

给定一个数组，求最长上升子序列。

思路

方法一：

dp[i] 为 以 A[i]结尾的LIS，那么对于dp[i]有dp[i] =max( dp[j] + 1) [ 0<= j < i, nums[j] < nums[i]  ]

效率为O(n^2)

方法二：

dp[i] 为 以 A[i]结尾的LIS ， g(i) = min ( A[j]) ( dp[j] = i) 即g(i)表示上升子序列为i，结尾最小的值。

比如，1，2，4，3中A[3] = 3

那么显然， g(1) <= g(2) <=……g(n)

我们可以用二分搜索查找满足g(k) >= A[i]的第一个下标k，则dp[i] = k ，此时 A[i] <= g(k)  ， 而dp[i] =k，所以更新g(k) = A[i]

python  O(n^2)

class Solution(object):
    def lengthOfLIS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums: return 0
        n = len(nums)
        dp = [1] * n
        for i in xrange(1, n):
            for j in xrange(i):
                if nums[i] > nums[j] and dp[i] < dp[j] + 1:
                    dp[i] = dp[j] + 1
        return max(dp)

python  O(nlogn)

class Solution(object):
    def lengthOfLIS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums: return 0

        n = len(nums)
        dp, g = [1] * n, [sys.maxint] * (n + 2)
        for i in xrange(n):
            k = self.lower_bound(1,n+1,nums[i],g)
            dp[i] = k
            g[k] = nums[i]
        return max(dp)

    def lower_bound(self,L, R, x,g):
        while L < R:
            mid = (L + R) >> 1
            if g[mid] < x:
                L = mid + 1
            else:
                R = mid
        return L