leetcode 674. Longest Continuous Increasing Subsequence

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.

Note: Length of the array will not exceed 10,000.

class Solution(object):
    def findLengthOfLCIS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums: return 0
        ans = 1
        cnt = 1
        for i in xrange(1, len(nums)):
            if nums[i] > nums[i-1]:
                cnt += 1
                ans = max(ans, cnt)
            else:
                cnt = 1
        return ans

本质上还是计数器！发现没有递增就reset计数器。

或者是用贪心也可以：

class Solution(object):
    def findLengthOfLCIS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """        
        ans = 0
        i = 0
        while i < len(nums):
            cnt = 1        
            while i+1<len(nums) and nums[i+1]>nums[i]:
                i += 1
                cnt += 1
            ans = max(ans, cnt)
            i += 1
        return ans

还有dp解法，虽然看起来不是那么舒服：

class Solution {
    public int findLengthOfLCIS(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        int n = nums.length;
        int[] dp = new int[n];
        
        int max = 1;
        dp[0] = 1;
        for (int i = 1; i < n; i++) {
            if (nums[i] > nums[i - 1]) {
                dp[i] = dp[i - 1] + 1;
            }
            else {
                dp[i] = 1;
            }
            max = Math.max(max, dp[i]);
        }
        
        return max;
    }
}