# 896.monotonic array

**896.Monotonic Array**

> 如果数组是单调递增或单调递减的，那么它是单调的。 如果对于所有 i <= j，A\[i] <= A\[j]，那么数组 A 是单调递增的。 如果对于所有 i <= j，A\[i]> = A\[j]，那么数组 A 是单调递减的。 当给定的数组 A 是单调数组时返回 true，否则返回 false。

```
示例 1：

输入：[1,2,2,3]
输出：true
示例 2：

输入：[6,5,4,4]
输出：true
示例 3：

输入：[1,3,2]
输出：false
示例 4：

输入：[1,2,4,5]
输出：true
示例 5：

输入：[1,1,1]
输出：true


提示：

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

方法： 题目比较简单，唯一的难点在于如何改进时间。最初想法如下，先从第一个开始判断，如果大于的往后判断，碰到小于的就返回false。小于也是一样，等于的话删掉第一个元素，递归调用。

```
class Solution {
public:
    bool isMonotonic(vector<int>& A) {
        int n=A.size();
        if(n==1)
            return true;
        if(A[0]>A[1] )
        {
            int i=2;
            while(i<n)
            {
                if(A[i-1]<A[i])
                    return false;
                i++;
            }
            return true;
        }
        if(A[0] < A[1])
        {
            int i=2;
            while(i<n)
            {
                if(A[i-1]>A[i])
                  return false;
                i++;
                   }
            return true;
        }
        if(A[0]==A[1])
        {
            A.erase(A.begin());
            return isMonotonic(A);
        }

           }
};
```

更短时间的例子如下，通过两个计数器遍历一遍向量即可。

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


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