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+// 81. Search in Rotated Sorted Array II
+#include "leetcode.h"
+
+/*
+ * there is an integer array 'nums' sorted in non-decreasing order (not
+ * necessarily with distinct values). before being passed to your function,
+ * 'nums' is rotated at an unknown pivot index 'k(0 <= 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
+ * '[0,1,2,4,4,4,5,6,6,7]' might be rotated at a pivot index 5 and become
+ * '[4,5,6,6,7,0,1,2,4,4]'. given the array 'nums' after the rotation and an
+ * integer 'target', return true if 'target' is in 'nums' of 'false' if it is
+ * not in 'nums'. you must decrease the overall operation steps as much as
+ * possible.
+ */
+
+class Solution {
+public:
+ bool search(vector<int> &nums, int target) {
+ int left = 0, right = nums.size() - 1;
+ while (left <= right) {
+ int mid = left + (right - left) / 2;
+ if (nums[mid] == target)
+ return true;
+ if (nums[left] == nums[mid] && nums[right] == nums[mid])
+ left++, right--;
+ else if (nums[left] <= nums[mid]) {
+ if (nums[left] <= target && nums[mid] > target)
+ right = mid - 1;
+ else
+ left = mid + 1;
+ } else {
+ if (nums[mid] < target && nums[right] >= target)
+ left = mid + 1;
+ else
+ right = mid - 1;
+ }
+ }
+ return false;
+ }
+};
+
+int main() {
+ Solution obj;
+ vector<int> nums = {2, 5, 6, 0, 0, 1, 2};
+ printf("%d\n", obj.search(nums, 0)); // expect: 1
+ printf("%d\n", obj.search(nums, 3)); // expect: 0
+}