Given an unsorted array of integers, find the length of longest continuous
increasing subsequence (subarray).
Example 1:
Input: [1,3,5,4,7] Output: 3 Explanation: The longest continuous increasing subsequence is [1,3,5], its length is 3. Even though [1,3,5,7] is also an increasing subsequence, it‘s not a continuous one where 5 and 7 are separated by 4.
Example 2:
Input: [2,2,2,2,2] Output: 1 Explanation: The longest continuous increasing subsequence is [2], its length is 1.
Idea 1. Dynamic programming, Let len[i] repents the length of the continuous increasing subsequence which ends at nums[i],
len[i] = len[i-1] + 1 if nums[i-1] < nums[i]
len[i] = 1 otherwise
Time complexity: O(N)
Space complexity: O(1)
class Solution { public int findLengthOfLCIS(int[] nums) { int len = 0; int maxLen = 0; for(int i = 0; i < nums.length; ++i) { if(i > 0 && nums[i-1] < nums[i]) { len += 1; } else { len = 1; } maxLen = Math.max(maxLen, len); } return maxLen; } }
Idea 1.b mark the anchor to store the start of the continuous increasing subsequence,
class Solution { public int findLengthOfLCIS(int[] nums) { int anchor = 0; int maxLen = 0; for(int i = 0; i < nums.length; ++i) { if(i > 0 && nums[i-1] >= nums[i]) { anchor = i; } maxLen = Math.max(maxLen, i - anchor + 1); } return maxLen; } }
原文地址:https://www.cnblogs.com/taste-it-own-it-love-it/p/10836936.html
时间: 2024-10-10 01:57:48