Longest Increasing Continuous Subsequence

Give an integer array，find the longest increasing continuous subsequence in this array.

An increasing continuous subsequence:

Can be from right to left or from left to right.
Indices of the integers in the subsequence should be continuous.

Notice

O(n) time and O(1) extra space.

Example

For [5, 4, 2, 1, 3], the LICS is [5, 4, 2, 1], return 4.

For [5, 1, 2, 3, 4], the LICS is [1, 2, 3, 4], return 4.

Runtime: 29ms

 1 class Solution {
 2 public:
 3     /**
 4      * @param A an array of Integer
 5      * @return  an integer
 6      */
 7     int longestIncreasingContinuousSubsequence(vector<int>& A) {
 8         // Write your code here
 9         
10         if (A.empty()) return 0;
11         int result = 1;
12         int leftMax = 1, rightMax = 1;
13         for (int i = 1; i < A.size(); i++) {
14             // from left to right
15             if (A[i] > A[i - 1]) {
16                 leftMax++;
17                 rightMax = 1;
18                 result = max(result, leftMax);
19             }
20             else { // from right to left
21                 leftMax = 1;
22                 rightMax++;
23                 result = max(result, rightMax);
24             }
25         }
26         return result;
27     }
28 };

Runtime: 838ms

 1 class Solution {
 2 public:
 3     /**
 4      * @param A an array of Integer
 5      * @return  an integer
 6      */
 7     int longestIncreasingContinuousSubsequence(vector<int>& A) {
 8         // Write your code here
 9         if (A.empty()) return 0;
10         
11         // scan from left to right
12         int leftMax = 1;
13         for (int i = 1; i < A.size(); i++) {
14             int j = i;
15             while (j < A.size() && A[j] > A[j - 1]) j++;
16             leftMax = max(leftMax, j - i + 1);
17         }
18         
19         // scan from right to left
20         int rightMax = 1;
21         for (int i = A.size() - 2; i >= 0; i--) {
22             int j = i;
23             while (j >= 0 && A[j] > A[j + 1]) j--;
24             rightMax = max(i - j + 1, rightMax);
25         }
26         return max(leftMax, rightMax);
27     }
28 };