描述
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.

問題描述

給定一個旋轉陣列，它是一個排序陣列通過在一個未知的位置經過旋轉而來，需要在該旋轉陣列中查詢一個目標值，如果存在返回它的索引，如果不存在，返回-1。

<span style="font-family:Microsoft YaHei;font-size:18px;">// 時間複雜度O(log n)，空間複雜度O(1)
class Solution {
public:
int search(int A[], int n, int target) {
int first = 0, last = n;
while (first != last) {
const int mid = (first + last) / 2;
if (A[mid] == target)
return mid;
if (A[first] <= A[mid]) {
if (A[first] <= target && target < A[mid])
last = mid;
else
first = mid + 1;
} else {
if (A[mid] < target && target <= A[last-1])
first = mid + 1;
else
last = mid;
}
}
return -1;
}
};</span>
 

JAVA版

<span style="font-family:Microsoft YaHei;font-size:18px;">public int search(int[] A, int target) {
        if(A==null||A.length==0)
                return -1;
        int first=0,last=A.length-1;
        while(first<=last){
                int mid=(first+last)/2;
                if(A[mid]==target){
                        return mid;
                }else if(A[mid]>=A[first]){
                        if(target>=A[first]&&target<A[mid]){
                                last=mid-1;
                        }else{
                                first=mid+1;
                        }
                }else{
                        if(target>A[mid]&&target<=A[last]){
                                first=mid+1;
                        }else{
                                last=mid-1;
                        }
                }
        }
        return -1;
    }
</span>