Search in Rotated Sorted Array

1.33. Search in Rotated Sorted Array

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

​​0 1 2 4 5 6 7​​ might become ​​4 5 6 7 0 1 2​​).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

class Solution {
public:
  int 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 mid;
      }
      else if (nums[mid]>=nums[left]){
        if (target > nums[mid]){
          left = mid + 1;
        }
        else{
          if (target >= nums[left]){
            right = mid - 1;
          }
          else{
            left = mid + 1;
          }
        }
      }
      else{
        if (target < nums[mid]){
          right = mid - 1;
        }
        else{
          if (target <= nums[right]){
            left = mid + 1;
          }
          else{
            right = mid - 1;
          }
        }
      }
    }
    return -1;
  }
};

2.81. Search in Rotated Sorted Array II

Follow up for "Search in Rotated Sorted Array":
What if duplicates

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

class Solution {
public:
  int 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;
      }
      else if (nums[mid]>nums[left]){
        if (target > nums[mid]){
          left = mid + 1;
        }
        else{
          if (target >= nums[left]){
            right = mid - 1;
          }
          else{
            left = mid + 1;
          }
        }
      }
      else if(nums[mid]<nums[left]){
        if (target < nums[mid]){
          right = mid - 1;
        }
        else{
          if (target <= nums[right]){
            left = mid + 1;
          }
          else{
            right = mid - 1;
          }
        }
      }
      else{
        left++;
      }
    }
    return false;
  }
};