[LeetCode] Longest Increasing Subsequence

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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.

这道题目,常规解法O(n2)没有难度,关键比较有意思的解题方式在于二分查找O(nlgn).

用一个数组tails,tails[i]的意思表示长度为i+1的LIS的末尾最小数。比如数组[2,3,5,4]长度为1的LIS最小的为2,那么tails[0]=2,长度为2的LIS有2,3和3,5末尾最小为3,所以tails[1]=3。以此类推有tails[2]=4。

反证可以确定,tails数组中的元素一定是有序递增的。所以可以在tails数组中进行二分查找:从前往后遍历输入的数组,查找大于当前数n的最小数的位置:如果找到了,那么表明该长度的LIS最小末尾该被替换;如果没有找到,则可以将n放置在原来最长LIS的后面组成一个长度加一的LIS。相当于动态生成了上面的tails数组。

代码如下:

public class Solution 
    public int lengthOfLIS(int[] nums) 
        int[] tails = new int[nums.length];
        int len = 0;
        for (int n : nums) 
            int l = 0,r = len;
            while (l != r) 
                int mid = (l + r) / 2;
                if (tails[mid] < n) l = mid + 1;
                else r = mid;
            
            tails[l] = n;
            if (l == len) len ++;
        
        return len;
    

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