LeetCode 674.Longest Continuous Increasing Subsequence-优快云博客

本文介绍了解决LeetCode 674题目的方法，通过一次遍历找到最长连续递增子序列的长度，并提供了一个简单易懂的C++实现示例。

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LeetCode 674.Longest Continuous Increasing Subsequence

Description:

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.

分析：

这道题其实就是动态规划找最长递增子序列的简易版，只要求找最长连续递增子串，可以减少一层循环，时间复杂度为O(n).

例子：
来看看例子[1,3,5,4,7]，首先我们用vector来定义一个length，表示子串的长度，先置为1，然后循环遍历这个vector容器，比较前后两个数的大小，更新每个length的大小。最后再取出length的最大值，即为结果。
首先长度置为1：length[0]=length[1]=length[2]=length[3]=length[4]=1
根据上述算法得到：
当nums[0]=1时, length[0]=1
当nums[1]=3时, 3>1, length[1]=2
当nums[2]=5时, 5>3, length[2]=3
当nums[3]=4时, 4<5, length[3]=1
当nums[4]=7时, 7>4, length[4]=2

代码如下：

#include <iostream>
#include <vector>
using namespace std;
class Solution {
public:
    int findLengthOfLCIS(vector<int>& nums) {
        vector<int> length(nums.size());
        for (int i = 0; i < nums.size(); i++) {
            length[i] = 1;
        }
        for (int i = 1; i < nums.size(); i++) {
            if (nums[i] > nums[i - 1]) {
                length[i] = length[i - 1] + 1;
            }
        }
        // 找出最大的length
        int max = 0;
        for (int i = 0; i < nums.size(); i++) {
            // cout << length[i] << " ";
            if (length[i] > max) max = length[i];
        }
        return max;
    }
};
int main() {
    Solution s;
    int n;
    cin >> n;
    vector<int> nums(n);
    for (int i = 0; i < n; i++) {
        cin >> nums[i];
    }
    cout << s.findLengthOfLCIS(nums) << endl;
    return 0;
}