标签：
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：1.在做完findMin之后，对做rotated很有启发。 2.以前写的花了40ms，具体什么方法还没回顾。 我终于懂了rotated是啥了。轮转暗暗
注意 findmin中while(a<b); rotated binary search中while(a<=b); Σ( ° △ °|||)︴介个好捉急。待我，想想怎么把while里面带不带=的问题解决掉。
class Solution { public: /*find the min, then use binary search to find target*/ int search(int A[], int n, int target) { int a=0,b=n-1,mid=0; while(a<b){ mid = (a+b)/2; if(A[mid]>A[b]) a=mid+1; else b=mid; } int indexOfMin = a; /*-----binary search----*/ int offset=indexOfMin; //if not rotated, min is the start element a=0;b=n-1; while(a<=b){ mid=(a+b)/2; int realMid=(mid+offset)%n; if(target == A[realMid]) return realMid; else if(target<A[realMid]) b=mid-1; else a=mid+1; } return -1; } };
算法里巧妙的地方在于 使用offset寻找realMid, target与A[realMid]比较，判断下一轮计算的realMid应该更小，还是更大。
Search in Rotated Sorted Array 【新思路】
标签：
原文地址：http://www.cnblogs.com/renrenbinbin/p/4332101.html
