题目要求:Search in Rotated Sorted Array
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.
代码如下:
class Solution { public: int search(int A[], int n, int target) { int index = -1; //题目简化成找到数组A中的元素位置…… for(int i = 0; i < n; i++){ if(A[i] == target){ index = i; break; } } return index; } };