[LeetCode] 674. Longest Continuous Increasing Subsequence 最長連續遞增序列

 

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.

 

這道題讓我們求一個數組的最長連續遞增序列，由於有了連續這個條件，跟之前那道 Number of Longest Increasing Subsequence 比起來，其實難度就降低了很多。可以使用一個計數器，如果遇到大的數字，計數器自增1；如果是一個小的數字，則計數器重置為1。用一個變量 cur 來表示前一個數字，初始化為整型最大值，當前遍歷到的數字 num 就和 cur 比較就行了，每次用 cnt 來更新結果 res，參見代碼如下：

 

解法一：

class Solution {
public:
    int findLengthOfLCIS(vector<int>& nums) {
        int res = 0, cnt = 0, cur = INT_MAX;
        for (int num : nums) {
            if (num > cur) ++cnt;
            else cnt = 1;
            res = max(res, cnt);
            cur = num;
        }
        return res;
    }
};

 

下面這種方法的思路和上面的解法一樣，每次都和前面一個數字來比較，注意處理無法取到錢一個數字的情況，參見代碼如下：

 

解法二：

class Solution {
public:
    int findLengthOfLCIS(vector<int>& nums) {
        int res = 0, cnt = 0, n = nums.size();
        for (int i = 0; i < n; ++i) {
            if (i == 0 || nums[i - 1] < nums[i]) res = max(res, ++cnt);
            else cnt = 1;
        }
        return res;
    }
};

 

Github 同步地址：

https://github.com/grandyang/leetcode/issues/674

 

類似題目：

Number of Longest Increasing Subsequence

Minimum Window Subsequence

 

參考資料：

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

https://leetcode.com/problems/longest-continuous-increasing-subsequence/discuss/107352/Java-code-6-liner

https://leetcode.com/problems/longest-continuous-increasing-subsequence/discuss/107365/JavaC%2B%2BClean-solution

 

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