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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.
分析:
1.mid的位置判断，若A[mid]>=A[left],则mid在左边区域；若A[mid]<A[left],则mid在右边区域
2.如果mid在左边区域，当A[left]<=target<A[mid],right=mid-1;
3.如果mid在右边区域，当A[mid]<target<=A[right],left=mid+1;
public class Solution {
public int search(int[] A, int target) {
int left=0;
int right=A.length-1;
while(right>=left) {
int mid=(right+left)/2;
if(A[mid]==target)
return mid;
if(A[mid]>=A[left]) {
if(target>=A[left]&&target<A[mid]) {
right=mid-1;
}else {
left=mid+1;
}
}else {
if(target>A[mid]&&target<=A[right]) {
left=mid+1;
}else {
right=mid-1;
}
}
}
return -1;
}
}
33-Search in Rotated Sorted Array
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原文地址：http://www.cnblogs.com/hwu2014/p/4423251.html
