Longest Continuous Increasing Subsequence LT674

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 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;
    }
} 

Longest Continuous Increasing Subsequence LT674

标签：sequence   star   math   example   time   ret   pac   mat   otherwise   

原文地址：https://www.cnblogs.com/taste-it-own-it-love-it/p/10836936.html