300. Longest Increasing Subsequence

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?难度：medium题目：给定一无序的整数数组，找出其递增的最大子序列。思路：动态规划， 用一个长度与输入数组相同长度的数组记录从0到当前元素的最大递增元素个数。dp(i) = Math.max(dp[j]…) + 1, (nums[i] > nums[j])dp(i) = 1 (nums[i] <= nums[j], 0 <= j <= i - 1)class Solution { public int lengthOfLIS(int[] nums) { int n = nums.length, maxIncLength = 0; int[] incLength = new int[n]; for (int i = 0; i < n; i++) { int curMaxIncLength = 0; for (int j = 0; j <= i; j++) { if (nums[j] < nums[i]) { curMaxIncLength = Math.max(curMaxIncLength, incLength[j]); } } incLength[i] = curMaxIncLength + 1; maxIncLength = Math.max(maxIncLength, incLength[i]); } return maxIncLength; }}