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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: nums = [10, 9, 8, 7],

Output: 10

我的想法

二分法呗。如果中点比后一个数小，说明在上坡，需要向后寻找，把开始指针移到中点位置；如果中点比前一个数小，说明在下坡，需要向前寻找，把结束指针移到中点位置

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

解答

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