leetcode_852

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.

Solutions

1. binary search

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