leetcode 674. Longest Continuous Increasing Subsequence（最长连续递增子序列）

寻找数组中最长连续递增子序列的算法解析

最新推荐文章于 2024-04-05 19:53:39 发布

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该博客介绍了如何找到一个未排序整数数组中长度最长的连续递增子序列。提供的Java代码实现通过遍历数组，判断相邻元素是否严格递增，动态维护子序列的长度。当遇到不满足递增条件的元素时，子序列长度重置为1。最终返回最大长度。示例展示了对于包含重复数字的情况，如何正确找到符合条件的子序列。

Given an unsorted array of integers nums, return the length of the longest continuous increasing subsequence (i.e. subarray). The subsequence must be strictly increasing.

A continuous increasing subsequence is defined by two indices l and r (l < r) such that it is [nums[l], nums[l + 1], …, nums[r - 1], nums[r]] and for each l <= i < r, nums[i] < nums[i + 1].

Example 1:

Input: nums = [1,3,5,4,7]
Output: 3
Explanation: The longest continuous increasing subsequence is [1,3,5] with length 3.
Even though [1,3,5,7] is an increasing subsequence, it is not continuous as elements 5 and 7 are separated by element
4.

Example 2:

Input: nums = [2,2,2,2,2]
Output: 1
Explanation: The longest continuous increasing subsequence is [2] with length 1. Note that it must be strictly
increasing.

找出最长的连续递增子序列，注意是连续的，而且是绝对递增的

思路：
遍历数组，如果满足绝对大于，count就加1，否则count重置为1

    public int findLengthOfLCIS(int[] nums) {
        if(nums == null || nums.length == 0) {
            return 0;
        }
        
        int count = 1;
        int result = 1;
        int n = nums.length;
        
        for(int i = 1; i < n; i++) {
            if(nums[i] > nums[i-1]) {
                count ++;
                result = Math.max(result, count);
            } else {
                count = 1;
            }
        }
        
        return result;
    }