标签：
Search in Rotated Sorted Array
问题：
Suppose a sorted array is rotated at some pivot unknown to you beforehand.
(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).
You are given a target value to search. If found in the array return its index, otherwise return -1.
You may assume no duplicate exists in the array.
思路：
二分查找
我的代码：
public class Solution { public int search(int[] A, int target) { if(A == null || A.length == 0) return -1; int left = 0; int right = A.length - 1; int index = helperSearch(A, left, right, target); return index; } public int helperSearch(int[] A, int left, int right, int target) { if(left == right) { if(A[left] == target) return left; else return -1; } if(left > right) return -1; int mid = (left + right)/2; if(A[mid] == target) return mid; if(A[left] < A[right]) { if(A[mid] > target) return helperSearch(A, left, mid - 1, target); else return helperSearch(A, mid + 1, right, target); } else { return Math.max(helperSearch(A, left, mid - 1, target), helperSearch(A, mid + 1, right, target)); } } }
他人代码：
public class Solution { public int search(int[] A, int target) { if(A == null || A.length == 0) return -1; int left = 0; int right = A.length - 1; while(left <= right) { int mid = (left + right)/2; if(A[mid] == target) return mid; if(A[left] <= A[mid]) { if(A[left] > target || A[mid] < target) left = mid + 1; else right = mid - 1; } else { if(A[right] < target || A[mid] > target) right = mid - 1; else left = mid + 1; } } return -1; } }
学习之处：
Search in Rotated Sorted Array
标签：
原文地址：http://www.cnblogs.com/sunshisonghit/p/4321426.html
