162. Find Peak Element

162. Find Peak Element

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.

方法一、二分查找

（1）非递归的写法

int findPeakElement(vector<int>& nums) {        int len = nums.size();        if(len <= 0)            return -1;                int low = 0;        int high = len - 1;        while(low < high)        {            int mid = low + (high - low) / 2;            int mid2 = mid + 1;            if(nums[mid] < nums[mid2])                low = mid2;            else                high = mid;        }        return low;    }

（2）递归的写法

int findPeakElement(vector<int>& nums) {        int len = nums.size();        if(len <= 0)            return -1;               return binaryRecusive(nums,0,len-1);    }        int binaryRecusive(vector<int>& nums,int low, int high)    {        if(low == high)            return low;        //while(low < high)        else{            int mid = low + (high - low) / 2;            int mid2 = mid + 1;            if(nums[mid] < nums[mid2])                return binaryRecusive(nums,mid2,high);            else                return binaryRecusive(nums,low,mid);        }        //return low;    }

方法二、顺序搜索

int findPeakElement(vector<int>& nums) {        for(int i = 1; i < nums.size(); i ++)        {            if(nums[i] < nums[i-1]) //思路是可行的，如果没有返回i-1,则表明前面的i个数字均满足，nums[i-1]<nums[i],也就是说未返回前面的数字中均满足第一个数字小于第二个数//则当再次遇到后面的一个数字小于前一个数字的，可以直接返回前一个数字的下标！！！            {                return i-1;            }        }        return nums.size()-1;    }