标签：
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; } }
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; } return -1; } }
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; } } };
学习之处：
标签：
原文地址：http://www.cnblogs.com/sunshisonghit/p/4319322.html
