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

思路：就是binary search；mid跟mid+1比较；

class Solution {
    public int peakIndexInMountainArray(int[] A) {
        if(A == null || A.length == 0) {
            return 0;
        }
        int start = 0; int end = A.length - 1;
        while(start + 1 < end) {
            int mid = start + (end - start) / 2;
            if(A[mid] > A[mid+1]) {
                end = mid;
            } else {
                start = mid;
            }
        }
        if(A[start] > A[end]) {
            return start;
        }
        return end;
    }
}

 

来源：CSDN

作者：flyatcmu

链接：https://blog.csdn.net/u013325815/article/details/104750924