线性表-Search in Rotated Sorted Arr

描述

Follow up for ”Search in Rotated Sorted Array”: What if duplicates are allowed?
Would this affect the run-time complexity? How and why?
Write a function to determine if a given target is in the array.

分析

此题允许重复元素，若按上题解法中A[m] >= A[l]，那么[l,m]为递增序列的假设就不成立了，如[2,3,2,2,2]。
如果A[m] >= A[l]不能确定递增，那么就进行拆分成两个条件：
若A[m] > A[l]，则区间[l,m]一定递增
若A[m] == A[l]，那么就l++,将重复元素略过。

代码实现 c++
class solution{
pubic: 
    bool search(const vector<int>& nums,int target){
        int first = 0,last = nums.size();
        while(first != last){
            const int mid = first + (last - first) / 2;
            if (nums[mid] == target)
                return true;
            if (nums[first] < nums[mid]){
                // 注意此处有等于号，表明在闭区间内
                if(nums[first] <= target && target < nums[mid])
                    last = mid;
                 else
                    first = mid + 1;
             } else if(nums[first] > nums[mid]){
                    if(nums[mid] < target && target <= nums[last - 1])
                        first = mid + 1;
                    else
                        last = mid;
              } else
                  //跳过重复元素
                  first ++;
          }
          return false;
    }
};

代码实现 python
class Solution:
    def search(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: bool
        """
        if not nums:
            return False
            
        start, end = 0, len(nums) - 1
        while start + 1 < end:
            mid = int(start + (end - start) / 2)
            if nums[mid] == target:
                return True
            elif nums[mid] > nums[start]:
                if target >= nums[start] and target <= nums[mid]:
                    end = mid
                else:
                    start = mid
            elif nums[mid] < nums[start]:
                if target >= nums[mid] and target <= nums[end]:
                    start = mid
                else:
                    end = mid
            else:
                start += 1
                    
        if nums[start] == target:
            return True
        if nums[end] == target:
            return True
        return False