Search in Rotated Sorted Array

Suppose a sorted array is rotated at some pivot unkonwn 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 values to search.If found in the array return its index,otherwise return -1.You may assume no duplicate exists in the array.

这个题目是用二分法在旋转数组中查找目标值。 旋转数组旋转后，原来总体有序的数组如今变成了分段有序。旋转轴左边的数组是有序的，旋转轴右边的一样是有序的，而且均是单调递增。 不过再使用二分法查找，可能会有些不一样。旋转轴可能不在mid的位置上，因此可能first~mid这段或者是mid~last这段并非递增有序的。

可是，若是知足a[mid]>=a[1],则在1~mid这段的元素就是递增的；

知足a[mid]< a[last],则在mid~last这段就是递增的。

解决方案：

class Solution
{
public:
    int search(const vector<int> &num,int target)
    {
        int first = 0;
        int last = num.size();
        while(first != last)
        {
            const int mid = first + (last - first)/2;
            if(num[mid] == target)
                return mid;
            if(num[first]<= num[mid])
            {
                if(num[first] <= target && num[mid] > target)
                {
                    last = mid;
                }
                else
                {
                    first = mid + 1;
                }
            }
            else
            {
                if(num[mid] < target && num[last-1] >= target)
                {
                    first = mid + 1;
                }
                else
                {
                    last = mid;
                }
            }
        }
        return -1;
    }
};

测试代码:

#include<iostream>
#include<vector>

using namespace std;


class Solution
{
public:
    int search(const vector<int>& num,int target)
    {
        int first = 0;
        int last = num.size();
        while(first != last)
        {
           const int mid = first + (last - first)/2;
           if(num[mid] == target)
               return  mid;
           if(num[first] <= num[mid])
           {
               if(num[first] <= target && num[mid] > target)
                   last = mid;
               else
                   first = mid + 1;
           }
           else
           {
               if(num[mid]<= target && num[last] > target)
                   first = mid + 1;
               else
                   last = mid;
           }
        }
        return -1;
    }
};

int main(void)
{

    vector<int> a;
    int target = 1;
    a.push_back(4);
    a.push_back(5);
    a.push_back(6);
    a.push_back(7);
    a.push_back(1);
    a.push_back(2);
    a.push_back(3);

    Solution s;
    int result = s.search(a,target);
    cout <<"The"<<target<<"is located in "<<result<<endl;
    return 0;
}