852. Peak Index in a Mountain Array

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Let‘s call an array A a mountain if the following properties hold:

• A.length >= 3
• There exists some 0 < i < A.length - 1 such that A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1]

Given an array that is definitely a mountain, return any i such that A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1].

Example 1:

Input: [0,1,0]
Output: 1

Input: [0,2,1,0]
Output: 1

Note:

1. 3 <= A.length <= 10000
2. 0 <= A[i] <= 10^6
3. A is a mountain, as defined above.
   因为山峰的数字肯定是大于左边的数字跟右边的数字，所以用二分查找左右逼近找速度比较快。
   
    

    1 int peakIndexInMountainArray(int* A, int ASize) {
    2     int low=0,high=ASize,mid;
    3     while(low<high){
    4         mid=low+(high-low)/2;
    5         if(A[mid]>A[mid+1]) high=mid;
    6         else    low=mid+1;
    7         
    8     }
    9     return low;
   10 }

    

852. Peak Index in a Mountain Array

标签：mon   span   hat   code   while   数字   NPU   HERE   array   

原文地址：https://www.cnblogs.com/real1587/p/9813923.html