leetcode 33. Search in Rotated Sorted Array

/* leetcode 33. Search in Rotated Sorted Array reference:http://blog.csdn.net/ljiabin/article/details/40453607 题目大意：一个排序数组绕某个数选择了，在这个旋转后的数组中找指定的元素target 找到就返回在数组中的序号，否则返回-1 解题思路： 1、当然是可以直接搜的（复杂度O(n)），也可以过.但是对于有一定顺序的数组，应该 有更高效的方法。 2、二分查找（复杂度O（logn））:二分查找时要注意转折点。其实只需要一边有序就可以。 分为3种情况 (1) 未旋转，如1，2，3，4，5，6. 条件是nums[left] < nums[right] (2) 旋转的位置超过了中点，如3，4，5，6，1，2。条件是nums[left]>nums[mid] (3) 旋转位置不超过中间点，如5，6，1，2，3，4。除了上面两种情况就是这一种了。 更新：3、(1)和(2)可以合并了写啊~这样两种情况，可以直接完成一次判断 I) nums[mid] < nums[right],说明mid到right为有序的，判断target是否在mid和right之间, 如果不在，那么一定在left和mid之间 II)nums[mid] >= nums[right],说明left到mid是有序的，判断target是否在left到mid之间， 如果不在，那么一定在left和mid之间 */

#include <iostream>
#include <vector>
using namespace std;

class Solution {
public:
    //考虑每种情况的
    int search(vector<int>& nums, int target) 
    {
        int len = nums.size();
        if (len == 0)
            return -1;
        return binarySearch(nums, 0, len - 1, target);
    }
    //1、对每一种情况进行分析，思路很好
    int binarySearch(vector<int>& nums, int left, int right, int target)
    {
        if (left > right)
            return -1;
        int mid = (left + right) / 2;
        if (nums[left] == target)
            return left;
        if (nums[mid] == target)
            return mid;
        if (nums[right] == target)
            return right;
        //(1)
        if (nums[left] <= nums[right])
        {
            if (target < nums[left] || target > nums[right])
                return -1;
            if (target < nums[mid] && target > nums[left])
                return binarySearch(nums, left + 1, mid - 1, target);
            else
                return binarySearch(nums, mid + 1, right - 1, target);
        }
        //(2)
        else if (nums[left] < nums[mid])
        {
            if (target > nums[left] && target < nums[mid])
                return binarySearch(nums, left + 1, mid - 1, target);
            else
                return binarySearch(nums, mid + 1, right - 1, target);
        }
        else //(3)
        {
            if (target > nums[mid] && target < nums[right])
                return binarySearch(nums, mid + 1, right - 1, target);
            else
                return binarySearch(nums, left + 1, mid - 1, target);
        }
    }

    //2、合起来写就是这样，不用递归快很多啊！
    int search2(vector<int>& nums, int target)
    {
        int left = 0;
        int right = nums.size() - 1;
        while (left <= right)
        {
            int mid = (left + right)/2;
            if (target == nums[mid])
                return mid;
            if (nums[left] <= nums[right]) //(1)
            {
                if (target < nums[mid])
                    right = mid - 1;
                else
                    left = mid + 1;
            }
            else if (nums[left] <= nums[mid]) //(2)
            {
                if (target > nums[mid] || target < nums[left])
                    left = mid + 1;
                else
                    right = mid - 1;
            }
            else
            {
                if (target < nums[mid] || target > nums[right])
                    right = mid - 1;
                else
                    left = mid + 1;
            }
        }

        return -1;
    }

    //3、把2中的合起来写就行了。因为只需要两种即可
    int search3(vector<int>& nums, int target)
    {
        int left = 0;
        int right = nums.size() - 1;
        while (left <= right)
        {
            int mid = (left + right) / 2;
            if (target == nums[mid])
                return mid;

            if (nums[mid] < nums[right])
            {
                if (target > nums[mid] && target <= nums[right])
                    left = mid + 1;
                else
                    right = mid - 1;
            }
            else
            {
                if (nums[left] <= target && nums[mid] > target)
                    right = mid - 1;
                else
                    left = mid + 1;
            }

        }

        return -1;
    }
};

void test_search()
{
    Solution sol;
    vector<int> nums1{278,280,281,286,287,290,2,3,4,8,9,14,15,16,21,24,25,31,32,34,36,37,42,45,51,52,54,55,60,63,66,68,69,71,76,81 };
    vector<int> nums2{ 7,8,1,2,3,4,5,6 };
    vector<int> nums3{ 3,4,5,6,7,8,1,2 };
    vector<int> nums{1};

    cout << sol.search(nums, 2) << endl;
    cout << sol.search(nums1, 286) << endl;
    cout << sol.search(nums2, 5) << endl;
    cout << sol.search(nums3, 5) << endl;

    cout << sol.search2(nums, 2) << endl;
    cout << sol.search2(nums1, 286) << endl;
    cout << sol.search2(nums2, 5) << endl;
    cout << sol.search2(nums3, 5) << endl;

    cout << sol.search3(nums, 2) << endl;
    cout << sol.search3(nums1, 286) << endl;
    cout << sol.search3(nums2, 5) << endl;
    cout << sol.search3(nums3, 5) << endl;
}


int main()
{
    test_search();
    return 0;
}