Longest Increasing Subsequence

yangrd 發布于2019-08-16 10:27 / 1358人閱讀

摘要：解題思路求不必連續的最長升序字符串長度，采用動態規劃，利用狀態表示以字符結尾的最長子串的長度，那么狀態轉移方程就是且必須小于另外還需維護一個最大長度。

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?

解題思路

求不必連續的最長升序字符串長度，采用動態規劃，利用狀態dp[i]表示以字符nums[i]結尾的最長子串的長度，那么狀態轉移方程就是:
dp[i]=Math.max(dp[i],dp[j]+1);0<=j且nums[j]必須小于nums[i]
另外還需維護一個最大長度。

2.代碼

public class Solution {
    public int lengthOfLIS(int[] nums) {
        if(nums.length==0) return 0;
        int[] dp=new int[nums.length];
        int maxLength=0;
        for(int i=0;i

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