题目描述
给出一个转动过的有序数组，你事先不知道该数组转动了多少
(例如,0 1 2 4 5 6 7可能变为4 5 6 7 0 1 2).
在数组中搜索给出的目标值，如果能在数组中找到，返回它的索引，否则返回-1。
假设数组中不存在重复项。
(例如,0 1 2 4 5 6 7可能变为4 5 6 7 0 1 2).
在数组中搜索给出的目标值，如果能在数组中找到，返回它的索引，否则返回-1。
假设数组中不存在重复项。
Suppose a sorted array is rotated at some pivot unknown to you beforehand.
(i.e.,0 1 2 4 5 6 7might become4 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.
示例1
输出
复制-1
class Solution {public: int search(int A[], int n, int target) { if(A == nullptr||n==0) return -1; int left = 0; int right = n-1; while(left<=right){ int mid = left + (right - left)/2; if(A[mid] == target) return mid; if(A[left] <= A[mid]){ if(A[left]<=target&&A[mid]>target) right = mid-1; else left = mid+1; } else{ if(A[mid]<target&&A[right]>=target) left = mid+1; else right = mid-1; } } return -1; }};