## Search in Rotated Sorted Array

• Time:O(\log n)
• Space:O(1)

## C++

``````class Solution {
public:
int search(vector<int>& nums, int target) {
int l = 0;
int r = nums.size() - 1;

while (l <= r) {
const int m = (l + r) / 2;
if (nums[m] == target)
return m;
if (nums[l] <= nums[m]) {  // nums[l..m] are sorted
if (nums[l] <= target && target < nums[m])
r = m - 1;
else
l = m + 1;
} else {  // nums[m..n - 1] are sorted
if (nums[m] < target && target <= nums[r])
l = m + 1;
else
r = m - 1;
}
}

return -1;
}
};
``````

## JAVA

``````class Solution {
public int search(int[] nums, int target) {
int l = 0;
int r = nums.length - 1;

while (l <= r) {
final int m = (l + r) / 2;
if (nums[m] == target)
return m;
if (nums[l] <= nums[m]) { // nums[l..m] are sorted
if (nums[l] <= target && target < nums[m])
r = m - 1;
else
l = m + 1;
} else { // nums[m..n - 1] are sorted
if (nums[m] < target && target <= nums[r])
l = m + 1;
else
r = m - 1;
}
}

return -1;
}
}
``````

## Python

``````class Solution:
def search(self, nums: List[int], target: int) -> int:
l = 0
r = len(nums) - 1

while l <= r:
m = (l + r) // 2
if nums[m] == target:
return m
if nums[l] <= nums[m]:  # nums[l..m] are sorted
if nums[l] <= target < nums[m]:
r = m - 1
else:
l = m + 1
else:  # nums[m..n - 1] are sorted
if nums[m] < target <= nums[r]:
l = m + 1
else:
r = m - 1

return -1
``````