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;
}
}