[LeetCode] 674. Longest Continuous Increasing Subsequence (C++)

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[LeetCode] 674. Longest Continuous Increasing Subsequence (C++)

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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.

class Solution {
public:
    int findLengthOfLCIS(vector<int>& nums) {
        int count=1; /* Why is "count" initialized to 1? You can simulate it on the draft paper */ 
        vector<int> recordLength; /* record the length of each continuous increasing subsequence (subarray) */
        if(nums.size()==0) {return 0;}
        if(nums.size()==1) {return 1;}
        if(nums.size()==2) {
            if(nums.at(1)>nums.at(0)) {
                return 2;
            }
            return 1;
        }
        for(int i=0;i<nums.size()-1;i++) {
            if(nums.at(i+1)>nums.at(i)) {
                count++;
            }
            else {
                recordLength.push_back(count);
                count=1;
            }
        }
        recordLength.push_back(count); /* do not forget to record the last continuous increasing subsequence's length */
        sort(recordLength.begin(),recordLength.end(),greater<int>()); 
        return recordLength.at(0); /* the length of longest continuous increasing subsequence (subarray) */
    }
};


/**
 * Submission Detail:
 * Runtime: 16 ms, faster than 98.05% of C++ online submissions for Longest Continuous Increasing Subsequence.
 * Memory Usage: 9.8 MB, less than 5.77% of C++ online submissions for Longest Continuous Increasing Subsequence.
 */