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There is an integer array which has the following features:
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.
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