300. 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.
Your algorithm should run in O(n2) complexity.
Follow up: Could you improve it to O(n log n) time complexity?
解题思路:定义一个dp数组,数组元素dp[i]表示以第i个元素结尾的递增序列的最大长度。DP算法,需要将前面元素的值都算出来public class Solution { public int lengthOfLIS(int[] nums) { if(nums.length == 0) return 0; int[] dp = new int[nums.length]; for(int i=0;i<nums.length;i++) { dp[i] = 1; } int max = 1; for(int i=1;i<nums.length;i++) { for(int j=0;j<i;j++) { if(nums[i]>nums[j] && dp[j]+1>dp[i]) { dp[i] = dp[j] +1; if(dp[i] > max) max=dp[i]; } } } return max; }}
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