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Longest Increasing Subsequence

yangrd / 1406人閱讀

摘要：解題思路求不必連續(xù)的最長(zhǎng)升序字符串長(zhǎng)度，采用動(dòng)態(tài)規(guī)劃，利用狀態(tài)表示以字符結(jié)尾的最長(zhǎng)子串的長(zhǎng)度，那么狀態(tài)轉(zhuǎn)移方程就是且必須小于另外還需維護(hù)一個(gè)最大長(zhǎng)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(xù)的最長(zhǎng)升序字符串長(zhǎng)度，采用動(dòng)態(tài)規(guī)劃，利用狀態(tài)dp[i]表示以字符nums[i]結(jié)尾的最長(zhǎng)子串的長(zhǎng)度，那么狀態(tài)轉(zhuǎn)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(hù)一個(gè)最大長(zhǎng)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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