【leetcode】Find Peak Element

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.
关键在于考虑一些边界的情况，如i=0和i=num.size()的情况

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

但是时间复杂度为O(log(n))的时候，需要用二分法

int findPeakElement(const vector<int> &num) {          int left=0,right=num.size()-1;          while(left<=right){              if(left==right)                  return left;              int mid=(left+right)/2;              if(num[mid]<num[mid+1])                  left=mid+1;              else                  right=mid;          }      }  

网上看到的另外一个好的方法
仔细分析下，不满足判断条件会导致i++, 也就是 num[i-1]>num[i-2]，再满足条件的话，就有num[i] < num[i-1]，因此i-1为峰值点。

int findPeakElement(const vector<int> &num) {          for(int i=1; i<num.size(); i++){              if(num[i] < num[i-1])                  return i-1;          }          return num.size()-1;      }