[LeetCode 33] Search in Rotated Sorted Array

Question

link

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.

Stats

Frequency	3
Difficulty	4
Adjusted Difficulty	4
Time to use	----------

Ratings/Color = 1(white) 2(lime) 3(yellow) 4/5(red)

Analysis

This question is not very difficult, yet commonly seen in interviews.

Without having any knowledge of pivot, we can check the mid-point value against left value and right value. Read this blog for more.

Solution

The code is easy to understand.

My code

Pay special attention to different larger/smaller conditions. It’s very easy to miss a equal sign or something.

public class Solution {
    public int search(int[] A, int target) {
        if (A == null || A.length == 0) {
            return -1;
        }
        int len = A.length;
        int left = 0;
        int right = len - 1;
        while (left + 1 < right) {
            int mid = left + (right - left) / 2;
            if (A[mid] == target) {
                return mid;
            } else if (A[left] < A[mid]) {
            // remember to pay attention to (A[left] == target) case
                if (A[left] <= target && target < A[mid]) {
                    right = mid;
                } else {
                    left = mid;
                }
            } else {
            // remember to pay attention to (A[right] == target) case
                if (A[mid] < target && target <= A[right]) {
                    left = mid;
                } else {
                    right = mid;
                }
            }
        }
        if (A[left] == target) {
            return left;
        } else if (A[right] == target) {
            return right;
        }
        return -1;
    }
}