标签:following write public least tco lin new array complex
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.
Given [1, 2, 1, 3, 4, 5, 7, 6]
Return index 1
(which is number 2) or 6
(which is number 7)
Time complexity O(logN)
halfhalf二分法(OOXXOOXX)。如果卡到爬坡知道右边肯定存在peak,如果卡到降坡知道左边肯定存在peak,所以每次还是可以果断改start, end取半的。
public 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){ throw new IllegalArgumentException(); } int start = 0; int end = A.length - 1; while (start + 1 < end){ int mid = start + (end - start) / 2; if (A[mid + 1] - A[mid] < 0){ end = mid; } else { start = mid; } } if (A[start] > A[end]){ return start; } return end; } }
lintcode75- Find Peak Element- medium
标签:following write public least tco lin new array complex
原文地址:http://www.cnblogs.com/jasminemzy/p/7580084.html