Leetcode-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.

Analysis:

Suppose we have three pointers:start, mid and end. There are three case for the array between start and end:

1. A[mid] < A[start]: Rotated, the maximum value is on the left of mid. If target < A[mid], then we search the left part; if target > A[mid], we then have two situations: (a) target <= A[end], we search the right part; (b) target > A[end], search the left part.

2. A[mid] > A[end]: Rotated, the maximum value is on the right of mid. target > A[mid]: search the left part; target < A[mid] && < A[start]: search the right part; target < A[mid] && >= A[start]: search the left part.

3. Otherwise: Not rotated. Just perform normal binary search.

Solution:

public class Solution {
public int search(int[] A, int target) {
if (A.length==0) return -1;

int start = 0, end = A.length-1;
int mid = -1;

while (start<=end){
mid = (start+end)/2;
if (A[mid]==target) return mid;

if (A[mid]<A[start]){
if (target<A[mid])
end = mid-1;
else if (target<=A[end])
start = mid+1;
else end = mid-1;
continue;
}

if (A[mid]>A[end]){
if (target>A[mid]) start = mid+1;
else if (target>=A[start]) end = mid-1;
else start = mid+1;
continue;
}

if (target<A[mid]) end = mid-1;
else start = mid+1;
}

return -1;
}
}