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 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, or false if it is not in nums.

You must decrease the overall operation steps as much as possible.

Example 1:

Input: nums = [2,5,6,0,0,1,2], target = 0
Output: true

Example 2:

Input: nums = [2,5,6,0,0,1,2], target = 3
Output: false

Constraints:

• 1 <= nums.length <= 5000
• -104 <= nums[i] <= 104
• nums is guaranteed to be rotated at some pivot.
• -104 <= target <= 104

Solution: Binary search or divide and conquer

If current range is ordered, use binary search, Otherwise, divide and conquer.

Time complexity: O(logn) best, O(n) worst
Space complexity: O(logn)

// Author: Huahua
class Solution {
public:
  bool search(vector<int>& A, int target) {
    return search(A, 0, A.size() - 1, target);
  }
private:
  bool search(vector<int>& A, int l, int r, int target) {
    if (l > r) return 0;
    if (l == r) return target == A[l];

    int mid = l + (r - l) / 2;

    if (A[l] < A[mid] && A[mid] < A[r])
      return (target <= A[mid]) ? search(A, l, mid, target) : search(A, mid + 1, r, target);
    else
      return search(A, l, mid, target) || search(A, mid + 1, r, target);
  }
};

The post 花花酱 LeetCode 81. Search in Rotated Sorted Array II appeared first on Huahua's Tech Road.

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
• -104 <= nums[i] <= 104
• All values of nums are unique.
• nums is an ascending array that is possibly rotated.
• -104 <= target <= 104

Solution: Binary Search

If the current range [l, r] is ordered, reduce to normal binary search. Otherwise, determine the range to search next by comparing target and nums[0].

Time complexity: O(logn)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  int search(vector<int>& nums, int target) {
    int l = 0;
    int r = nums.size();
    while (l < r) {
      int m = l + (r - l) / 2;
      int x = (nums[m] < nums[0]) == (target < nums[0])
              ? nums[m]
              : target < nums[0] ? INT_MIN : INT_MAX;
      if (x < target)
        l = m + 1;
      else if (x > target)
        r = m;
      else
        return m;
    }
    return -1;
  }
};

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