Find Peak Element

Find Peak Element

问题：

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.

思路：

　　二分查找

我的思路：

public class Solution {

    public int findPeakElement(int[] num) {

        if(num == null || num.length == 0) return -1;

        int len = num.length;

        if(len == 1 || num[0] > num[1]) return 0;

        if(num[len - 1] > num[len - 2]) return len - 1;

        int left = 1;

        int right = len - 2;

        return getPeakIndex(num, left, right);

        

    }

    public int getPeakIndex(int[] num, int left, int right)

    {

        if(left > right) return -1;

        if(left == right)

        {

            if(num[left] > num[left - 1] && num[left] > num[left + 1]) return left;

            else return -1;

        }

        int mid = (left + right)/2;

        if(num[mid] > num[mid - 1] && num[mid] > num[mid + 1])  return mid;

        int index = getPeakIndex(num, left, mid - 1);

        if(index == -1)

            index = getPeakIndex(num, mid + 1, right);

        return index;

    }

    

    

}

View Code

改进后算法：

public class Solution {

    public int findPeakElement(int[] num) {

        if(num == null || num.length == 0) return -1;

        int len = num.length;

        if(len == 1) return 0;

        int left = 0;

        int right = len - 1;

        while(left <= right)

        {

            int mid = (left + right)/2;

            if(mid == 0)

            {

                if(num[mid] > num[mid+1])   return mid;

            }

            else if(mid == len - 1)

            {

                if(num[mid] > num[mid-1])   return mid;

            }

            else

            {

                if(num[mid] > num[mid-1] && num[mid] > num[mid+1]) return mid;

            }

            if(num[mid] < num[mid+1]) left = mid + 1;

            else right = mid - 1 ;

        }

        return -1;

        

    }

}

View Code

他人代码：

class Solution {

public:

    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;

        }

    }

};

View Code

回来用模板写的代码：

    public int findPeak(int[] A) {

        int[] num = A;

        if(num == null || num.length == 0) return -1;

        int len = num.length;

        if(len == 1) return 0;

        int left = 0;

        int right = len - 1;

        while(left + 1 < right)

        {

            int mid = (left + right)/2;

            if(mid == 0)

            {

                if(num[mid] > num[mid+1])   return mid;

            }

            else if(mid == len - 1)

            {

                if(num[mid] > num[mid-1])   return mid;

            }

            else

            {

                if(num[mid] > num[mid-1] && num[mid] > num[mid+1]) return mid;

                if(num[mid] < num[mid+1]) left = mid + 1;

                else right = mid ;

            }

        }

        return num[left]>num[right] ? left:right;

    }

View Code

 

学习之处：

• 首先肯定存在peak点，因为最左侧num[-1]=num[n]=负无穷大，肯定存在一个Peak
• 若中间点不是peak，则分别朝着左右两侧上升的地方走
• 左右left和right，都往着上升的地方走，相遇的地方就是peak点（这个有点难理解啊），没证明出来，但是一想是这么一回事