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;    }};