LeetCode 300. Longest Increasing Subsequence

分析

难度 中
来源 https://leetcode.com/problems/longest-increasing-subsequence/
思路
tails is an array storing the smallest tail of all increasing subsequences with length i+1 in tails[i].
For example, say we have nums = [4,5,6,3], then all the available increasing subsequences are:

len = 1 : [4], [5], [6], [3] => tails[0] = 3
len = 2 : [4, 5], [5, 6] => tails[1] = 5
len = 3 : [4, 5, 6] => tails[2] = 6

We can easily prove that tails is a increasing array. Therefore it is possible to do a binary search in tails array to find the one needs update.

Each time we only do one of the two:

(1) if x is larger than all tails, append it, increase the size by 1
(2) if tails[i-1] < x <= tails[i], update tails[i]

题目

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

解答

Runtime: 0 ms, faster than 100.00% of Java online submissions for Longest Increasing Subsequence.
Memory Usage: 34.3 MB, less than 89.04% of Java online submissions for Longest Increasing Subsequence.

package LeetCode;

public class L300_LongestIncresingSubsequence {
    public int lengthOfLIS(int[] nums) {
        if(nums.length<1)
            return 0;
        int[] tails=new int[nums.length+1];
        tails[0]=nums[0];
        int len=1;

        for(int i=1;itails[len-1]){
                tails[++len-1]=nums[i];
            }else{
                for(int j=0;jtails[j]&&nums[i]