Leetoced 81. Search in Rotated Sorted Array II

Follow up for "Search in Rotated Sorted Array":What if duplicates are allowed?Would this affect the run-time complexity? How and why?

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.

class Solution {public:    typedef vector<int>::iterator It;    bool search(vector<int>& nums, int target) {        It begin = nums.begin();        It end = nums.end();        return reSearch(begin, end, target);    }private:    bool reSearch(It begin, It end, int target) {        size_t n = end - begin;        if (n == 0) return false;        if (n == 1) return target == *begin;        It mid = begin + n / 2;        if (*mid == target) return true;        if (*mid == *begin)  // 1, 3, 1, 1 // 当mid == begin 会 bug            return (reSearch(begin + 1, mid, target) || reSearch(mid + 1, end, target));        else if (*mid > *begin ) {            if (target == *begin) return true;            else if (target > *begin && target < *mid) return reSearch(begin, mid, target);            else                 return reSearch(mid + 1, end, target);        }         else { // *mid < *begin            if (target == *(end - 1)) return true;            else if (target > *mid && target < *(end - 1)) return reSearch(mid + 1, end, target);            else                 return reSearch(begin, mid, target);        }            return false;    }};

参考后

class Solution {public:    bool search(vector<int>& nums, int target) {        if (nums.empty()) return false;        int b = 0;        int e = nums.size() - 1;        while (b <= e) {            int mid = (b + e) / 2;            if (nums[mid] == target)                return true;            if (nums[mid] == nums[b])                ++b;            else if (nums[mid] < nums[b]) {                if (nums[e] == target) return true;                if (nums[mid] < target && nums[e] > target) {                    b = mid + 1;                    --e;                } else {                    e = mid - 1;                }            } else {                if (nums[b] == target) return true;                else if (nums[b] < target && target < nums[mid]) {                    ++b;                    e = mid - 1;                } else {                    b = mid + 1;                }            }        }        return false;    }};