[leetcode]162. Find Peak Element

题目链接：https://leetcode.com/problems/find-peak-element/#/description

A peak element is an element that is greater than its neighbors.

Given an input array where num[i] ≠ num[i+1], find a peak element and return its index.

The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.

You may imagine that num[-1] = num[n] = -∞.

For example, in array [1, 2, 3, 1], 3 is a peak element and your function should return the index number 2.

click to show spoilers.

Note:

Your solution should be in logarithmic complexity.

方法一：（顺序查找）

class Solution {
public:
    int findPeakElement(vector& nums) {
        int n = nums.size();
        if(n == 1)
            return 0;
        if(nums[0] > nums[1])
            return 0;
        if(nums[n-1] > nums[n-2])
            return n-1;
        for(int i = 1; i < n-1; i ++)
            if(nums[i] > nums[i-1] && nums[i] > nums[i+1])
                return i;
    }
};  

方法二：（递归二分查找）

class Solution {
public:
    int findPeakElement(vector& nums) {
        return Helper(nums, 0, nums.size()-1);
    }
    int Helper(vector& nums, int low, int high)
    {
        if(low == high)
            return low;
        int mid = low + (high-low)/2;
        if(nums[mid] > nums[mid+1])
            return Helper(nums, low, mid);
        else
            return Helper(nums, mid+1, high);
    }
};  

方法三：（迭代二分查找）

class Solution {
public:
    int findPeakElement(vector& nums) {
        int low = 0;
        int high = nums.size()-1;
        while(low < high)
        {
            int mid = low + (high-low)/2;
            if(nums[mid] > nums[mid+1])
                high = mid;
            else
                low = mid+1;
        }
        return low;
    }
};