LeetCode/33.search-in-rotated-sorted-array.cpp at master · samba9274/LeetCode · GitHub

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/*
 * @lc app=leetcode id=33 lang=cpp
 *
 * [33] Search in Rotated Sorted Array
 *
 * https://leetcode.com/problems/search-in-rotated-sorted-array/description/
 *
 * algorithms
 * Medium (38.22%)
 * Likes:    15978
 * Dislikes: 988
 * Total Accepted:    1.6M
 * Total Submissions: 4.2M
 * Testcase Example:  '[4,5,6,7,0,1,2]\n0'
 *
 * 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
 * -10^4 <= nums[i] <= 10^4
 * All values of nums are unique.
 * nums is an ascending array that is possibly rotated.
 * -10^4 <= target <= 10^4
 *
 *
 */

// @lc code=start
class Solution
{
public:
    int search(vector<int> &nums, int target)
    {
        unsigned short int minIndex = 0, minNum, i = 0, mid, l, h;
        while (i < nums.size())
        {
            if (nums[i] < minNum)
            {
                minIndex = i;
                minNum = nums[i];
            }
            i++;
        }
        rotate(nums.begin(), nums.begin() + minIndex, nums.end());
        for (mid = nums.size() / 2, l = 0, h = nums.size(); l < h && nums[mid] != target;)
        {
            if (nums[mid] < target)
                l = mid + 1;
            else
                h = mid;
            mid = (l + h) / 2;
        }
        if (l >= h)
            return -1;
        return (mid + minIndex) % nums.size();
    }
};
// @lc code=end