852. Peak Index in a Mountain Ar

描述

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.

法一：暴力搜索

class Solution {
public:
    int peakIndexInMountainArray(vector<int>& A) {
        int maxNum = A[0];
        int index = 0;
        for(int i = 1; i < A.size(); i++)
        {
            if(maxNum<A[i])
            {
                index = i;
                maxNum=A[i];
            }
        }
        
        return index;
        
    }
};

法二：寻找A[i]>A[i+1]的地方

class Solution {
public:
    int peakIndexInMountainArray(vector<int>& A) {
        for(int i = 0;i+1<A.size();i++)
        {
            if(A[i]>A[i+1])
                return i;
        }
        
    }
};

法三：二分查找

int peakIndexInMountainArray(vector<int> A) {
        int l = 0, r = A.size() - 1, mid;
        while (l < r) {
            mid = (l + r) / 2;
            if (A[mid] < A[mid + 1]) l = mid;
            else if (A[mid - 1] > A[mid]) r = mid;
            else return mid;
        }
    }