Suppose an array sorted in ascending order 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.
Your algorithm's runtime complexity must be in the order of O(log n).
Example 1:
Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4
Example 2:Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1
这道题的核心是用二分法,间或应用一些其他的技巧,可以说是二分法的进阶版。。。
class Solution {
public:
int search(vector<int>& nums, int target) {
int left = 0;
int right = nums.size() - 1;
while(left <= right)
{
int mid = left + (right - left)/2;
if(nums[mid] == target) return mid;
if(nums[mid] < nums[right])
{
if(target > nums[mid] && target <= nums[right]) left = mid + 1;
else right = mid - 1;
}
else
{
if(target > nums[mid] || target <= nums[right]) left = mid + 1;
else right = mid - 1;
}
}
return -1;
}
};