300. Longest Increasing Subsequence

题目：

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?

链接： http://leetcode.com/problems/longest-increasing-subsequence/

题解：

求数组的最长递增子序列。经典dp问题，在很多大学讲DP的教程里，都会出现这道题以及Longest Common Subsequence。 这里其实也有O(nlogn)的方法，比如Patience Sorting一类的，二刷再研究。下面我们来看DP。这个问题一开始可以被分解为recursive的子问题，一步一步优化就可以得到带有memorization的iterative解法。初始化dp[i] = 1，即一个元素的递增序列。 假设以i - 1结尾的subarray里的LIS为dp[i - 1]，那么我们要求以i结尾的subarray里的LIS，dp[i]的时候，要把这个新的元素和之前所有的元素进行比较，同时逐步比较dp[j] + 1与dp[i]，假如发现更长的序列，我们则更新dp[i] = dp[j] + 1，继续增加j进行比较。当i之前的元素全部便利完毕以后，我们得到了当前以i结尾的subarray里的LIS，就是dp[i]。

Time Complexity - O(n²)， Space Complexity - O(n²)。

public class Solution {
    public int lengthOfLIS(int[] nums) {
        if(nums == null || nums.length == 0) {
            return 0;
        }
        int len = nums.length, max = 0;
        int[] dp = new int[len];

        for(int i = 0; i < len; i++) {
            dp[i] = 1;
            for(int j = 0; j < i; j++) {
                if(nums[i] > nums[j] && dp[j] + 1 > dp[i]) {
                    dp[i] = dp[j] + 1;
                }
            }
            max = Math.max(max, dp[i]);
        }

        return max;
    }
}

题外话：

#300题！又是一个里程碑了。虽然之前做的很多题目都忘记了，但相信二刷会好好巩固和再学习。微信群里一起刷题的小伙伴们，好多已经拿到了Amazon的Offer，我也要好好努力才行啊。这周休假在家，周三继续修理房子，希望一切顺利。 同时希望在这周能够把LeetCode第一遍完成，然后早日学习新的知识，比如多线程，设计模式，以及一些系统设计等等。

Reference:

https://leetcode.com/discuss/67609/short-java-solution-using-dp-o-n-log-n

https://leetcode.com/discuss/67554/9-lines-c-code-with-o-nlogn-complexity

https://leetcode.com/discuss/67533/c-typical-dp-2-solution-and-nlogn-solution-from-geekforgeek

https://leetcode.com/discuss/67565/simple-java-o-nlogn-solution

https://leetcode.com/discuss/71129/space-log-time-short-solution-without-additional-memory-java

https://leetcode.com/discuss/67687/c-o-nlogn-solution-with-explainations-4ms

https://leetcode.com/discuss/69309/c-o-nlogn-with-explanation-and-references

https://leetcode.com/discuss/67572/o-nlogn-and-o-n-2-java-solutions

https://leetcode.com/discuss/67689/4ms-o-nlogn-non-recursive-easy-to-understand-java-solution

https://leetcode.com/discuss/67553/share-java-dp-solution

https://leetcode.com/discuss/72127/easy-to-understand-solution-using-dp-with-video-explanation

https://leetcode.com/discuss/67806/another-o-n-log-n-python

http://www.geeksforgeeks.org/longest-monotonically-increasing-subsequence-size-n-log-n/

http://www.cs.cornell.edu/~wdtseng/icpc/notes/dp2.pdf

https://courses.engr.illinois.edu/cs473/sp2011/lectures/08_notes.pdf

http://www.cs.toronto.edu/~vassos/teaching/c73/handouts/lis.pdf

http://www.cs.mun.ca/~kol/courses/2711-w08/dynprog-2711.pdf

https://courses.cs.washington.edu/courses/cse417/02wi/slides/06dp-lis.pdf

https://www.cs.princeton.edu/courses/archive/spring13/cos423/lectures/LongestIncreasingSubsequence.pdf

https://en.wikipedia.org/wiki/Patience_sorting

https://en.wikipedia.org/wiki/Longest_increasing_subsequence