[LeetCode] 674. Longest Continuous Increasing Subsequence 最长连续递增序列
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.
给一个没有排序的整数数组，找出最长的连续递增子序列（子数组）。
Java:
public int findLengthOfLCIS(int[] nums) {
int res = 0, cnt = 0;
for(int i = 0; i < nums.length; i++){
if(i == 0 || nums[i-1] < nums[i]) res = Math.max(res, ++cnt);
else cnt = 1;
}
return res;
}
Python:
class Solution(object):
def findLengthOfLCIS(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
result, count = 0, 0
for i in xrange(len(nums)):
if i == 0 or nums[i-1] < nums[i]:
count += 1
result = max(result, count)
else:
count = 1
return result
Python: wo
class Solution(object):
def findLengthOfLCIS(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if not nums:
return 0
n = len(nums)
dp = [0] * n
dp[0] = 1
longest = 1
for i in xrange(1, n):
if nums[i] > nums[i-1]:
dp[i] = dp[i-1] + 1
else:
dp[i] = 1
longest = max(longest, dp[i])
return longest
C++:
int findLengthOfLCIS(vector<int>& nums) {
int res = 0, cnt = 0;
for(int i = 0; i < nums.size(); i++){
if(i == 0 || nums[i-1] < nums[i]) res = max(res, ++cnt);
else cnt = 1;
}
return res;
}
类似题目：
[LeetCode] 300. Longest Increasing Subsequence 最长递增子序列
[LeetCode] 673. Number of Longest Increasing Subsequence 最长递增序列的个数
