// 33. Search in Rotated Sorted Array #include "leetcode.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 */ class Solution { public: int search(vector &nums, int target) { int left = 0, right = nums.size() - 1; while (left < right) { int mid = (left + right) / 2; if ((nums[0] > target) ^ (nums[0] > nums[mid]) ^ (target > nums[mid])) left = mid + 1; else right = mid; } return left == right && nums[left] == target ? left : -1; } }; int main() { Solution obj; vector n1 = {4, 5, 6, 7, 0, 1, 2}, n2 = {4, 5, 6, 7, 0, 1, 2}, n3 = {1}; printf("%d\n", obj.search(n1, 0)); // expect: 4 printf("%d\n", obj.search(n2, 3)); // expect: -1 printf("%d\n", obj.search(n3, 0)); // expect: -1 }