Search in Rotated Sorted Array

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.

二分查找的推广，需要增加判断条件，为方便举例我们假设要找的数字为3，先用mid把数列分为两种情况，再讨论两种情况的子情况

mid = first+ ( end - first)/2;

if ( A[mid ] == target ) return mid;(找到终止)

if ( A[first] <=  A[mid])   (这个时候还得考虑两种情况，1 2 3 4 5 6 7 8 9 和 5 6 7 8 9 1 2 3 4)

{

if( A[first] < target && target < =A[mid])

{ end=mid ;}

else

{ first= mid+1;}

}

else   ./A[first] > A[mid]) (7 8 9 1 2 3 4 5 6  和  9 1 2 3 4 5 6 7 8)

{

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

{ first = mid+1}

else

{ end = mid;}

}

class Solution {public:    int search(int A[], int n, int target) {        int first=0;        int end=n;        while(first!=end)        {            int mid =first+ (end - first)/2;            if(A[mid] == target) return mid;            if(A[first] <= A[mid])            {                if(A[first] <= target && target<A[mid] )                {                    end = mid;                }                else                {                    first = mid+1;                }            }            else            {                if(A[mid] < target && target <= A[end-1] )                {                    first = mid+1;                }                else                {                    end = mid;                }                            }        }        return -1;    }};