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

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.

Approach #1: Binary Search.

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

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

852. Peak Index in a Mountain Array

原文：https://www.cnblogs.com/ruruozhenhao/p/9937235.html