LeetCode-Longest Continuous Increasing Subsequence

最新推荐文章于 2020-06-09 11:18:11 发布

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最新推荐文章于 2020-06-09 11:18:11 发布

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分类专栏： LeetCode 文章标签： LeetCode Java

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博客围绕给定的一维数组，要求找出最长连续增长的子数组。通过定义起始位置，找到连续增长的最后一个位置得出子数组长度，再比较各长度得到结果。给出了示例说明，且数组长度不超10000。

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Description:
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.

题意：给定一个一维数组，要求找到最长连续增长的子数组；

解法：定义一个起始位置，找到连续增长的最后一个位置，就可以得出这个子数组的长度，比较各个长度得到其最长连续增长的子数组的长度；

class Solution {
    public int findLengthOfLCIS(int[] nums) {
        int st = 0;
        int maxLen = 0;
        for(int i = 1; i < nums.length; i++) {
            if (nums[i] <= nums[i-1]) {
                maxLen = Math.max(maxLen, i - st);
                st = i;
            }
        }
        maxLen = Math.max(maxLen, nums.length - st);
        return maxLen;
    }
}