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.
给定一个有序的反转过的数组。这个数组的第一个值比最后一个值要大。
所给数组中没有重复元素。
class Solution {
public:
int search(int A[], int n, int target) {
if(n==0)return -1;
if(A[0]<A[n-1]){
//数组未发生反转，正常的二分查找
int low=0, high=n-1;
while(low<=high){
int mid=(low+high)/2;
if(A[mid]==target)return mid;
else if(target<A[mid])high=mid-1;
else low=mid+1;
}
}
else{
//数组未发生反转
int low=0, high=n-1;
while(low<=high){
int mid=(low+high)/2;
if(A[mid]==target)return mid;
else if(target>A[mid]){
//先判断的位置，在确定target的位置
if(A[mid]>=A[0])low=mid+1;
else if(A[mid]<=A[n-1]){
if(target<=A[n-1])low=mid+1;
else if(target>=A[0])high=mid-1;
else break;
}
else break;
}
else{
//先判断的位置，在确定target的位置
if(A[mid]<=A[n-1])high=mid-1;
else if(A[mid]>=A[0]){
if(target>=A[0])high=mid-1;
else if(target<=A[n-1])low=mid+1;
else break;
}
else break;
}
}
}
return -1;
}
};LeetCode: Search in Rotated Sorted Array [032],布布扣,bubuko.com
LeetCode: Search in Rotated Sorted Array [032]
原文地址：http://blog.csdn.net/harryhuang1990/article/details/26222199
