LeetCode-in-All
33. Search in Rotated Sorted Array
Medium
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
Solution
#include <stdio.h>
int search(int* nums, int numsSize, int target) {
int lo = 0;
int hi = numsSize - 1;
int mid;
while (lo <= hi) {
mid = ((hi - lo) >> 1) + lo;
if (nums[mid] == target) {
return mid;
}
// Check if the left half is sorted
if (nums[lo] <= nums[mid]) {
// Target is in the left half
if (nums[lo] <= target && target <= nums[mid]) {
hi = mid - 1;
} else { // Target is in the right half
lo = mid + 1;
}
}
// Otherwise, the right half is sorted
else {
// Target is in the right half
if (nums[mid] <= target && target <= nums[hi]) {
lo = mid + 1;
} else { // Target is in the left half
hi = mid - 1;
}
}
}
return -1;
}