# 852. Peak Index in a Mountain Array

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:

1. 3 <= A.length <= 10000
2. 0 <= A[i] <= 10^6
3. A is a mountain, as defined above.

class Solution {
public:
int peakIndexInMountainArray(vector<int>& A) {
return max_element(A.begin(), A.end()) - A.begin();
}
};

class Solution {
public:
int peakIndexInMountainArray(vector<int>& A) {
for (int i = 1; i < (int)A.size() - 1; ++i) {
if (A[i] > A[i + 1]) return i;
}
return 0;
}
};

class Solution {
public:
int peakIndexInMountainArray(vector<int>& A) {
int n = A.size(), left = 0, right = n - 1;
while (left < right) {
int mid = left + (right - left) / 2;
if (A[mid] < A[mid + 1]) left = mid + 1;
else right = mid;
}
return right;
}
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/852

Find Peak Element

https://leetcode.com/problems/peak-index-in-a-mountain-array/

https://leetcode.com/problems/peak-index-in-a-mountain-array/discuss/139848/C%2B%2BJavaPython-Better-than-Binary-Search

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