LWC 49：674. Longest Continuous Increasing Subsequence

LWC 49：674. Longest Continuous Increasing Subsequence

传送门：674. Longest Continuous Increasing Subsequence

从今天开始改变下刷题策略，Leetcode Weekly Contest中的每一题都单独更新，有些题还不再自己的能力范围之内，暂时不去解答，有能力独立AC后，再写解题报告。

Problem:

Given an unsorted array of integers, find the length of longest continuous increasing subsequence.

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.

思路：
用了DP记录当前序列的最大连续长度。

dp[i] 表示当前位置的最大连续长度更新：dp[i] = dp[i - 1] + 1 if (前一元素小于当前元素)

代码如下：

    public int findLengthOfLCIS(int[] nums) {        if (nums.length == 0) return 0;        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] = dp[i - 1] + 1;            }            max = Math.max(max, dp[i]);        }        return max;    }

此题不一定需要DP，直接用单个变量控制即可。

代码如下：

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