Search in Rotated Sorted Array 62

Question

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.

Example

For [4, 5, 1, 2, 3] and target=1, return 2.

For [4, 5, 1, 2, 3] and target=0, return -1.

Challenge

O(logN) time

Solution

如果没有重复元素，就是二分搜索，mid要么落在大元素的区间，要么落在小元素的区间。先看mid落在前半区间还是后半区间，再看target是在该区间的前半部分还是后半部分。

代码如下：

No duplicate

public class Solution {
    /** 
     *@param A : an integer rotated sorted array
     *@param target :  an integer to be searched
     *return : an integer
     */
    public int search(int[] A, int target) {
        // write your code here
        if (A == null || A.length == 0){
            return -1;
        }

        int start = 0;
        int end = A.length - 1;
        while (start + 1 < end){
            int mid = start + (end - start)/2;
            if (A[mid] == target){
                return mid;
            }
            if (A[start] < A[mid]){
                if (A[start] <= target && target <= A[mid]){
                    end = mid;
                }else{
                    start = mid;
                }
            }else{
                if (A[mid] <= target && target <= A[end]){
                    start = mid;
                }else{
                    end = mid;
                }
            }
        }

        if (A[start] == target){
            return start;
        }else if (A[end] == target){
            return end;
        }else{
            return -1;
        }
    }
}

With duplicate

public class Solution {
    /** 
     * param A : an integer ratated sorted array and duplicates are allowed
     * param target :  an integer to be search
     * return : a boolean 
     */
    public boolean search(int[] A, int target) {
        // write your code here
        if(A == null || A.length == 0){
            return false;
        }

        for(int i = 0; i < A.length; i++){
            if(A[i] == target){
                return true;
            }
        }

        return false;
    }
}