Leetcode: Find Peak Element

标签：leetcode   binary search   

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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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;
    }

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;
        }
    }

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

Leetcode: Find Peak Element

标签：leetcode   binary search   

原文地址：http://blog.csdn.net/doc_sgl/article/details/41991809