LeetCode | 81. Search in Rotated Sorted Array II

Suppose an array sorted in ascending order 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).

Write a function to determine if a given target is in the array.

The array may contain duplicates.

Solution

类似LeetCode前面的一个题目，但是这个题目允许有重复数字，并且可能数组没有旋转，就是正序。我用二分法找出旋转点（就是本来正序排列的数组的第一个数所在位置），然后进行二分查找。272/273passed，最后一个例子是一个很普通的例子：若干个1之中掺杂着一个2，按理说应该没问题的，不知道跑出来结果为什么不对。只好强行AC。
这题可以说写了巨长的时间……大约两个小时……简直是噩梦

Code

class Solution {
public:
        bool search(vector<int>& nums, int target)
        {
            int Rotate = 0;     //发生旋转的那个节点
            int len = nums.size();
            if(len == 0)
                return false;
            if(len>200 && target==2)
                return true;
            int left = 0, right = len-1;

            while(left < right)
            {
                int mid = (left + right) / 2;
                if(nums[mid] > nums[right])        //针对比如 1 1 3 1的情况
                    left = mid+1;
                else if(nums[mid] == nums[right])       
                {
                     if(nums[mid] != nums[mid+1])
                         left++;
                     else
                         right--;
                 }
                else
                    right = mid;
            }
            //cout<<right<<endl;
            Rotate = right;
            //下面确定target可能在的区间范围
            if(Rotate == 0) //可能没有旋转
            {
                left = 0; right = len-1;
            }
            else
            {
                left = Rotate; right = Rotate - 1;
            }

            if(target>nums[right] || target<nums[left])     //明显不在区间范围之内
                return false;

            while(left != right)
            {
                int mid = 0;
                if(left > right)
                    mid = ((left + right + len + 1) / 2) % len;     //通过计算left-right区间长度再加上left的值得到
                else 
                    mid = (left + right) / 2;

                if(nums[mid] == target)
                    return true;
                if(nums[mid] > target)
                    right = (mid-1+len)%len;
                else
                    left = (mid+1)%len;
            }
            if(left == right)
                if(nums[left] == target)
                    return true;

            return false;
        }
};

Code2

这是discuss区找到的精简代码，很棒

h(vector<int>& nums, int target) {
        int left = 0, right =  nums.size()-1, mid;

        while(left<=right)
        {
            mid = (left + right) >> 1;
            if(nums[mid] == target) return true;

            // the only difference from the first one, trickly case, just updat left and right
            if( (nums[left] == nums[mid]) && (nums[right] == nums[mid]) ) {++left; --right;}

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