Problem
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.
Solution
we scan from left to right, a sequence start at i ==0 or a[i] <= a[i-1], the start of next sequence is also the end of the previous sequence.
public class IncreasingSequence {
public static int longest(int[] a) {
//1,3,5,4,7
int max = 0;
int count = 1;
for(int i = 1; i < a.length; i++) { //4
if(a[i] > a[i-1]) {
count++; //2
} else {
max = Math.max(count, max);
count = 1;
}
}
max = Math.max(count, max);
return max;
}
public static void main(String...args) {
int[] a = new int[]{1,3,5,4,7};
System.out.println(longest(a));
}
}
public static int longest(int[] a) {
//1,3,5,4,7
int max = 0;
int count = 1;
for(int i = 1; i < a.length; i++) { //4
if(a[i] > a[i-1]) {
count++; //2
} else {
max = Math.max(count, max);
count = 1;
}
}
max = Math.max(count, max);
return max;
}
public static void main(String...args) {
int[] a = new int[]{1,3,5,4,7};
System.out.println(longest(a));
}
}
The time cost is O(N), the space cost is O(1)
