Leetcode: 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.

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Note:

Your solution should be in logarithmic complexity.

解法一：O(n)，比较次数2n

int findPeakElement(const vector<int> &num) {//stupid O(n), compare 2n times.
        int sz = num.size();
		for(int index = 0; index < sz; index++)
		{
			bool isleft = true;
			bool isright = true;
			if(index > 0) isleft = num[index-1] < num[index] ? true : false;
			if(index <= sz-2) isright = num[index] > num[index+1] ? true : false;
			if(isleft && isright) return index;
		}
		return -1;
    }

解法二：O(n)，比较次数n

int findPeakElement(const vector<int> &num) {//smart O(n), compare n times.
        for(int i=1; i<num.size(); i++){
            if(num[i] < num[i-1])
                return i-1;
        }
        return num.size()-1;
    }

为什么呢？因为按照题意，num[0]是大于左边的不存在的那个元素的，num[size-1]也是大于右边那个不存在的元素的，假如不存在，那么就会有num[0]<num[1]，num[1]<num[2]，就是增序，num[size-2]<num[size-1]，这样num[size-1]就是peak elem了，所以一定存在。

于是就是这样的思路，num[NULL] < num[0],我们假设左边的元素小于右边的元素，那么第一个左边元素大于右边的那个一定是peak elem.如num[0].为什么第一个就是呢？因为前面的都是左<右，判断左>右为false。

解法三：O(logN)

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;
        }
    }

和解法二同样的道理。

参考http://blog.csdn.net/u010367506/article/details/41943309