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+// 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<int> &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<int> 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
+}