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lintcode-medium-Find Peak Element

时间:2016-03-21 08:12:37      阅读:134      评论:0      收藏:0      [点我收藏+]

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There is an integer array which has the following features:

  • The numbers in adjacent positions are different.
  • A[0] < A[1] && A[A.length - 2] > A[A.length - 1].

We define a position P is a peek if:

A[P] > A[P-1] && A[P] > A[P+1]

Find a peak element in this array. Return the index of the peak.

 

Given [1, 2, 1, 3, 4, 5, 7, 6]

Return index 1 (which is number 2) or 6 (which is number 7)

 

class Solution {
    /**
     * @param A: An integers array.
     * @return: return any of peek positions.
     */
    public int findPeak(int[] A) {
        // write your code here
        if(A == null || A.length == 0)
            return 0;
        
        int left = 1;
        int right = A.length - 2;
        
        while(left < right - 1){
            
            int mid = left + (right - left) / 2;
        
            if(A[left] <= A[mid] && A[mid] <= A[right])
                left = mid;
            else if(A[left] >= A[mid] && A[mid] >= A[right])
                right = mid;
            else if(A[left] <= A[mid] && A[right] >= A[mid])
                right--;
            else
                left++;
            
        }
        
        if(A[left] > A[right])
            return left;
        else
            return right;
    }
}

 

lintcode-medium-Find Peak Element

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原文地址:http://www.cnblogs.com/goblinengineer/p/5300475.html

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