Leetcode_search-in-rotated-sorted-array

http://oj.leetcode.com/problems/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：先遍历找到pivot，那么就是两个有序的序列。判断target在哪一块，再做二分查找。

参考代码：

class Solution {public:    int bs(int A[], int left, int right, int target)    {        if(left>right)            return -1;        int mid = (left+right)/2;        if(A[mid]<target)        {            left=mid+1;            return bs(A, left, right, target);        }        else if(A[mid]>target)        {            right = mid-1;            return bs(A, left, right, target);        }        return mid;    }    int search(int A[], int n, int target) {        int i = 0;        while(i+1<n && A[i]<A[i+1])        {            ++i;        }        if(A[0]<=target && i< n && target<=A[i])        {            return bs(A, 0, i, target);        }        else if(i+1 < n && A[i+1]<=target && target<=A[n-1])        {            return bs(A, i+1, n-1, target);        }        return -1;    }};

//SECOND TRIAL, O(logn) time

class Solution {private:    int findInRotatedArray(int A[], int left, int right, int target)    {        if(!A || left > right)            return -1;        if(left==right)        {            if(A[left]==target)                return left;            else                return -1;        }        int mid = (left+right)/2;        if(A[left]<A[mid])        {            if(A[left]<=target && target<=A[mid])                return findInRotatedArray(A, left, mid, target);            else                return findInRotatedArray(A, mid+1, right, target);        }        else if(A[left]>A[mid])        {            if(target>=A[mid] && target<=A[right])                return findInRotatedArray(A, mid, right, target);            else                return findInRotatedArray(A, left, mid-1, target);        }        else        {            if(A[mid]==A[right])                return A[left]==target ? left : -1;            int ans = findInRotatedArray(A, left, mid, target);            if(ans<0)                return findInRotatedArray(A, mid+1, right, target);            return ans;        }    }public:    int search(int A[], int n, int target) {        return findInRotatedArray(A, 0, n-1, target);    }};