Given an unsorted array of integers, find the length of longest continuous increasing subsequence (subarray).
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| 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.
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| Example 2:
Input: [2,2,2,2,2]
Output: 1
Explanation: The longest continuous increasing subsequence is [2], its length is 1.
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Note: Length of the array will not exceed 10,000.
Solution:
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| class Solution {
public int findLengthOfLCIS(int[] nums) {
int ans = 0;
int anchor = 0;
for (int i = 0; i < nums.length; i++) {
if (i > 0 && nums[i - 1] >= nums[i]) anchor = i;
ans = Math.max(ans, i - anchor + 1);
System.out.printf("i %d anchor %d %n", i, anchor);
}
return ans;
}
}
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Solution 2021-10-26#
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| class Solution {
public int findLengthOfLCIS(int[] nums) {
int[] dp = new int[nums.length];
Arrays.fill(dp, 1);
int max = 1;
for (int i = 1; i < nums.length; i++) {
if (nums[i] > nums[i - 1]) {
dp[i] = Math.max(dp[i - 1] + 1, dp[i]);
}
max = Math.max(dp[i], max);
}
return max;
}
}
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