leetcode problem 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.

 

 

思路：

　　给你一个已排序好但移位了的数组，找到特定的数。画一幅图就很容易理解了。一个被旋转的有序数组A[1..n]，假定转折点是A[k]，那么A[k+1] < A[k+2] < ... < A[n] < A[1] < A[2] < ... < A[k]

　　可以用二分查找稍微改点型：虽然我们不知道转折点k在哪，但是我们还是可以通过比较A[mid]与A[start],A[end]来确定要找的目标书target是在A[start,...,mid]中，还是在A[mid+1,...,end]  所以时间复杂度还是lg(n)

 

代码：

Runtime: 12 ms

class Solution{

public:

    int search(int A[], int n, int target) {

        if (n <= 0)

            return -1;

        if (n == 1)

            return *A == target ? 0 : -1;



        int begin = 0, end = n, mid = (begin + end) / 2;



        if (A[begin] <= A[mid-1]) {

            if (target >= A[begin] && target <= A[mid-1]) {

                auto it = lower_bound(A, A + mid, target);

                if (*it == target)

                    return it - A;

                else 

                    return -1;

            }

            int res = search(A + mid, end - mid, target);

            return res == -1 ? -1 : mid + res;

        }

        else {

            if (target >= A[mid] && target <= A[end-1]) {

                auto it = lower_bound(A+mid, A+end, target);

                if (*it == target)

                    return it - A;

                else

                    return -1;

            }    

            int res = search(A + begin, mid - begin, target);

            return res == -1 ? -1 : begin + res;

        }

    }

};