题目:
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.
思路分析:
(1)如果left<right,直接用二分搜索
(2)if left>right
则 mid=(left+right)>>1;
if (A[mid])==target 则停止
else find(A,left,mid-1,target)与find(A,mid+1,right,target).
源代码:
class Solution { public: int search(int A[], int n, int target) { return find(A,0,n-1,target); } private: int find(int A[],int left,int right,int target){ if(left>right) return -1; int idx=-1; if(A[left]<=A[right]){ while(left<=right){ int mid=(left+right)>>1; if(A[mid]==target){ idx=mid; break; } else if(A[mid]>target) right=mid-1; else left=mid+1; } } else{ int mid=(left+right)>>1; if(A[mid]==target) idx=mid; else{ idx=find(A,left,mid-1,target); idx=((idx==-1)?find(A,mid+1,right,target):idx); } } return idx; } };
LeetCode-Search in Rotated Sorted Array
原文：http://www.cnblogs.com/sixue/p/4086886.html