Leetcode刷题75-896. 单调数列(C++详细解法！！！)

最新推荐文章于 2024-03-24 23:01:37 发布

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最新推荐文章于 2024-03-24 23:01:37 发布

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分类专栏： LeetCode从零开始文章标签： LeetCode C++

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Come from : [https://leetcode-cn.com/problems/monotonic-array/]

1.Question

An array is monotonic if it is either monotone increasing or monotone decreasing.

An array A is monotone increasing if for all i <= j, A[i] <= A[j]. An array A is monotone decreasing if for all i <= j, A[i] >= A[j].

Return true if and only if the given array A is monotonic.

Example 1 :

Input: [1,2,2,3]
Output: true

Example 2 :

Input: [6,5,4,4]
Output: true

Example 3 :

Input: [1,3,2]
Output: false

Example 4 :

Input: [1,2,4,5]
Output: true

Example 5 :

Input: [1,1,1]
Output: true

Note :

1. 1 <= A.length <= 50000
2. -100000 <= A[i] <= 100000

2.Answer

easy 类型题目。
这题我走了好多弯路。。。
感觉自己好蠢啊。。。

class Solution {
public:
    bool isMonotonic(vector<int>& A) {
        int tag = 0;            //标志位
        for(int i = 0; i < A.size() - 1; ++i)
        {
            if(A[i] == A[i+1])
            {
                continue;
            }
            if(A[i] > A[i+1])   //判断是否单调减  设置tag == 1
            {
                if(tag == 2)
                {
                    return false;
                }
                tag = 1; 
                continue;
            }
            if(A[i] < A[i+1])  //判断是否单调增  设置tag == 2
            {
                if(tag == 1)
                {
                    return false;
                }
                tag = 2;
                continue;
            }
        }
        return true;
    }
};

3.大神们的解决方案

方法（有点两个辅助值的意思）：很简单，不多BB。

class Solution {
public:
    bool isMonotonic(vector<int>& A) {
        int up = 1, down = 1;            
        for(int i = 0; i < A.size() - 1; ++i)
        {
            if(A[i] >= A[i+1])
            {
                ++down;
            }
            if(A[i] <= A[i+1])   
            {
                ++up;
            }
        }
        if(up == A.size() || down == A.size())
        {
            return true;
        }
        else
        {
            return false;
        }
       
    }
};