33. Search in Rotated Sorted Array && 81. Search in Rotated Sorted Array II &&

33. Search in Rotated Sorted Array

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

(i.e., 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.

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Binary Search Array

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(M) Search in Rotated Sorted Array II (M) Find Minimum in Rotated Sorted Array

public class Solution {
  public int search(int[] nums, int target) {
    int len = nums.length;
    int leftMin = nums[0];
    int rightMax = nums[len - 1];

    if (target > rightMax && target < leftMin)
      return -1;
    boolean targetAtLeft = target >= leftMin;

    int left = 0;
    int right = len - 1;

    while (left <= right) {
      int mid = (left + right) / 2;
      int midVal = nums[mid];
      if (midVal == target)
        return mid;
      if (targetAtLeft) {
        if (midVal >= leftMin && midVal < target)
          left = mid + 1; // go right
        else
          right = mid - 1; //go left
      } else {
        if (midVal <= rightMax && midVal > target)
          right = mid - 1; // go left
        else
          left = mid + 1; //go right
      }
    }
    return -1;
  }
}

81. Search in Rotated Sorted Array II

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

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

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

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Array Binary Search

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(H) Search in Rotated Sorted Array