[LintCode] 62 Search in Rotated Sorted Array 解题报告

Description
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

思路
用二分法，判断target的区间。
如果A[mid] > A[start]并且target在它们之间，往前找。否则往后找。注意后面还是一个Sorted Array！
如果A[mid] < A[end]并且target在它们之间，往后找。否则往前找。注意前面还是一个Sorted Array！

Code

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