LeetCode 674. Longest Continuous Increasing Subsequence

[Woodyiiiiiii]

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.

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这是一道简单的线性规划问题。题目要求的是找到最长的连续增长子数组。那么我们可以设置一个一维线性DP数组，dp[i]表示以nums[i]结尾的最长连续递增子数组。状态转移方程是dp[i] = (nums[i] > nums[i - 1]) ? dp[i - 1] + 1 : 1。

class Solution {
    public int findLengthOfLCIS(int[] nums) {
        int n = nums.length;
        if (n == 0) return 0;
        int[] dp = new int[n];
        dp[0] = 1;
        int maxLength = 1;
        for (int i = 1; i < n; ++i) {
            dp[i] = (nums[i] > nums[i - 1]) ? dp[i - 1] + 1 : 1;
            maxLength = Math.max(maxLength, dp[i]);
        }
        return maxLength;
    }
}

可以继续优化，将空间复杂度降至为常数级别：

class Solution {
    public int findLengthOfLCIS(int[] nums) {
        int n = nums.length;
        if (n == 0) return 0;
        int maxLength = 1, preSum = 1;
        for (int i = 1; i < n; ++i) {
            int sum = (nums[i] > nums[i - 1]) ? preSum + 1 : 1;
            maxLength = Math.max(maxLength, sum);
            preSum = sum;
        }
        return maxLength;
    }
}

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参考资料:

• https://leetcode.com/problems/longest-continuous-increasing-subsequence/submissions/