852. Peak Index in a Mountain Array

Let's call an array arr a mountain if the following properties hold:

• arr.length >= 3
• There exists some i with 0 < i < arr.length - 1 such that:
  • arr[0] < arr[1] < ... arr[i-1] < arr[i]
  • arr[i] > arr[i+1] > ... > arr[arr.length - 1]

Given an integer array arr that is guaranteed to be a mountain, return any i such that arr[0] < arr[1] < ... arr[i - 1] < arr[i] > arr[i + 1] > ... > arr[arr.length - 1].

Example 1:

Input: arr = [0,1,0]
Output: 1

Example 2:

Input: arr = [0,2,1,0]
Output: 1

Example 3:

Input: arr = [0,10,5,2]
Output: 1

Example 4:

Input: arr = [3,4,5,1]
Output: 2

Example 5:

Input: arr = [24,69,100,99,79,78,67,36,26,19]
Output: 2

Constraints:

• 3 <= arr.length <= 104
• 0 <= arr[i] <= 106
• arr is guaranteed to be a mountain array.

Follow up: Finding the O(n) is straightforward, could you find an O(log(n)) solution?

class Solution {
    public int peakIndexInMountainArray(int[] arr) {
        for(int i = 1; i < arr.length - 1; i++) {
            if(arr[i] > arr[i - 1] && arr[i] > arr[i + 1]) return i;
        }
        return -1;
    }
}

O(n)就直接写，O（logn）就二分法。

class Solution {
    public int peakIndexInMountainArray(int[] arr) {
        int l = 0, r = arr.length - 1;
        while(l < r) {
            int m = l + (r - l) / 2;
            if(arr[m] < arr[m + 1]) l = m + 1;
            else r = m;
        }
        return l;
    }
}

判断的不再是m和l、r，而是m和m+1，比如1,2,3,4,1,

如果m 《  m+1，说明l 到 m 都是递增的，peak一定在m右边，l = m + 1.

否则r = m。

最后返回的是l。