Search in Rotated Sorted Array – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
There is an integer array nums sorted in ascending order (with distinct values).
Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= 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 possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.
You must write an algorithm with O(log n) runtime complexity.
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
numsare unique. numsis an ascending array that is possibly rotated.-104 <= target <= 104
C++ Search in Rotated Sorted Array LeetCode Solution
class Solution {
public:
int search(vector<int>& nums, int target) {
int l = 0, r = nums.size()-1;
while (l <= r) {
int mid = (l+r) / 2;
if (target == nums[mid])
return mid;
// there exists rotation; the middle element is in the left part of the array
if (nums[mid] > nums[r]) {
if (target < nums[mid] && target >= nums[l])
r = mid - 1;
else
l = mid + 1;
}
// there exists rotation; the middle element is in the right part of the array
else if (nums[mid] < nums[l]) {
if (target > nums[mid] && target <= nums[r])
l = mid + 1;
else
r = mid - 1;
}
// there is no rotation; just like normal binary search
else {
if (target < nums[mid])
r = mid - 1;
else
l = mid + 1;
}
}
return -1;
}
};
Java Search in Rotated Sorted Array LeetCode Solution
public class Solution {
public int search(int[] nums, int target) {
int start = 0;
int end = nums.length - 1;
while (start <= end){
int mid = (start + end) / 2;
if (nums[mid] == target)
return mid;
if (nums[start] <= nums[mid]){
if (target < nums[mid] && target >= nums[start])
end = mid - 1;
else
start = mid + 1;
}
if (nums[mid] <= nums[end]){
if (target > nums[mid] && target <= nums[end])
start = mid + 1;
else
end = mid - 1;
}
}
return -1;
}
}
Python 3 Search in Rotated Sorted Array LeetCode Solution
class Solution:
# @param {integer[]} numss
# @param {integer} target
# @return {integer}
def search(self, nums, target):
if not nums:
return -1
low, high = 0, len(nums) - 1
while low <= high:
mid = (low + high) / 2
if target == nums[mid]:
return mid
if nums[low] <= nums[mid]:
if nums[low] <= target <= nums[mid]:
high = mid - 1
else:
low = mid + 1
else:
if nums[mid] <= target <= nums[high]:
low = mid + 1
else:
high = mid - 1
return -1
Array-1180
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