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 thatA[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
Example 2:
Input: [0,2,1,0]
Output: 1
Note:
3 <= A.length <= 10000
0 <= A[i] <= 10^6
- 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 }