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Search in Rotated Sorted Array.cpp
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Search in Rotated Sorted Array.cpp
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/*
Search in Rotated Sorted Array
==============================
There is an integer array nums sorted in ascending order (with distinct values).
Prior to being passed to your function, nums is rotated at an unknown pivot index k (0 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].
Given the array nums after the rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.
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
Example 3:
Input: nums = [1], target = 0
Output: -1
Constraints:
1 <= nums.length <= 5000
-104 <= nums[i] <= 104
All values of nums are unique.
nums is guaranteed to be rotated at some pivot.
-104 <= target <= 104
Follow up: Can you achieve this in O(log n) time complexity?
*/
class Solution
{
public:
int search(vector<int> &nums, int target)
{
int i = 0, j = nums.size() - 1, mid = -1;
while (i <= j)
{
mid = (i + j) / 2;
if (target == nums[mid])
return mid;
if (nums[i] <= nums[mid])
{
if (target <= nums[mid] && target >= nums[i])
j = mid - 1;
else
i = mid + 1;
}
else
{
if (target >= nums[mid] && target <= nums[j])
i = mid + 1;
else
j = mid - 1;
}
}
return -1;
}
};