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.

Note:

Your solution should be in logarithmic complexity.

摘自：https://leetcode.com/discuss/18107/o-logn-solution-javacode

This problem is similar to Local Minimum. And according to the given condition, num[i] != num[i+1], there must exist a O(logN) solution. So we use binary search for this problem.

• If num[i-1] < num[i] > num[i+1], then num[i] is peak
• If num[i-1] < num[i] < num[i+1], then num[i+1...n-1] must contains a peak
• If num[i-1] > num[i] > num[i+1], then num[0...i-1] must contains a peak
• If num[i-1] > num[i] < num[i+1], then both sides have peak (n is num.length)

public int findPeakElement(int[] num) 
    {    
        return helper(num,0,num.length-1);
    }

    public int helper(int[] num,int start,int end)
    {
        if(start == end)
            return start;
        else if(start+1 == end)
        {
            if(num[start] > num[end]) 
                return start;
            return end;
        }
        else
        {
    
            int m = (start+end)/2;
    
            if(num[m] > num[m-1] && num[m] > num[m+1]){
    
                return m;
    
            }else if(num[m-1] > num[m] && num[m] > num[m+1]){
    
                return helper(num,start,m-1);
    
            }else{
    
                return helper(num,m+1,end);
    
            }
        }
    }