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// 33. Search in Rotated Sorted Array
#include <stdio.h>
#include <stdlib.h>
/*
* 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 tht the resulting
* array is '[nums[k], nums[k + 1], ..., nums[n - 1], nums[0], nums[1], ...,
* nums[k - 1]]' (0-indxed). 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 otherwise. you must write an algorithm with 'O(log n)'
* runtime complexity
*/
int search(int *nums, int nums_size, int target) {
int left, right;
for (left = 0, right = nums_size - 1; left <= right;) {
int mid = (left + right) / 2;
if (nums[mid] == target)
return mid;
if (nums[left] <= nums[mid]) {
if (nums[left] <= target && target < nums[mid])
right = mid - 1;
else
left = mid + 1;
} else {
if (nums[mid] < target && target <= nums[right])
left = mid + 1;
else
right = mid - 1;
}
}
return -1;
}
int main() {
int n1[] = {4, 5, 6, 7, 0, 1, 2}, n2[] = {4, 5, 6, 7, 0, 1, 2}, n3[] = {1};
printf("%d\n", search(n1, 7, 0)); // expect: 4
printf("%d\n", search(n2, 7, 3)); // expect: -1
printf("%d\n", search(n3, 1, 0)); // expect: -1
}
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