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
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.
核心思路:二分查找
代码:
int peakIndexInMountainArray(vector<int>& A) {
int left=0,right=A.size()-1;
int res=left+(right-left)/2;
while(left<=right){
res=left+(right-left)/2;
if(A[res]<A[res-1]){
right=res-1;
}else if(A[res]<A[res+1]){
left=res+1;
}else{
break;
}
}
return res;
}