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