Find Peak Element(ARRAY - Devide-and-Conquer)

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QUESTION

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.

1st TRY

时间复杂度要达到O(logn)必须用分治法。

如果array[mid-1]大于array[mid]，则左边的子数组array[start..mid-1]肯定有peak element（因为array[start]总是大于左边的元素）；同样地，如果array[mid+1]大于array[mid]，则由边的子数组array[mid+1..end]肯定有peak element（因为array[end]总是大于右边的元素）。

class Solution {
public:
    int findPeakElement(const vector<int> &num) {
        return binarySearch(num, 0, num.size()-1);
    }
    int binarySearch(const vector<int> &num, int start, int end)
    {
        if(end - start <= 1)
        {
            if(num[start]>num[end]) return start;
            else return end;
        }
        
        int mid = (start+end) >> 1;
        if(num[mid] > num[mid+1]) return binarySearch(num, start, mid);
        else return binarySearch(num, mid+1, end);
    }
};

Result: Accepted

Find Peak Element(ARRAY - Devide-and-Conquer)

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原文地址：http://www.cnblogs.com/qionglouyuyu/p/4220105.html