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  • 81. Search in Rotated Sorted Array II

    题目描述:

    Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
    
    (i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]).
    
    You are given a target value to search. If found in the array return true, otherwise return false.
    
    Example 1:
    
    Input: nums = [2,5,6,0,0,1,2], target = 0
    Output: true
    Example 2:
    
    Input: nums = [2,5,6,0,0,1,2], target = 3
    Output: false
    Follow up:
    
    This is a follow up problem to Search in Rotated Sorted Array, where nums may contain duplicates.
    Would this affect the run-time complexity? How and why?

    解题思路:这题与33. Search in Rotated Sorted Array之间的区别在于数组中可能会有重复的元素。解题思路与前者一样,关键仍然是找到单调递增区域。在前者的代码中,nums[first] <= nums[mid]这句就不适用了,应将”=“去掉。因为数组的旋转点可能会位于重复的元素中间,这样当”=“条件成立时,first与mid之间并不一定是单调递增的,有可能是重复的元素被截断所致,因此只需要first++跳过重复的元素重新从第一步开始即可。但”<“条件成立时,一定是单调递增的。

    参考代码:

    #include <vector>
    #include <iostream>
    
    using namespace std;
    
    
    class Solution
    {
     public:
      bool search(const vector<int>& nums, int target)
      {
        int first = 0, last = nums.size() - 1;
        while (first != last + 1)
        {
          const int mid = first + (last - first) / 2;
          if (nums[mid] == target)
          {
            return true;
    //        return mid;
          }
          if (nums[first] < nums[mid])
          {
            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])
            {
              first = mid + 1;
            }
            else
            {
              last = mid;
            }
          }
          else
          {
            first++;
          }
        }
        return false;
    //    return -1;
      }
    };
    
    
    int main()
    {
        int a[] = {2, 2, 5, 6, 0, 1, 2};
        Solution solu;
        vector<int> vec_arr(a, a+7);
        bool index = solu.search(vec_arr, 2);
        cout << "target index: " << index << endl;
    
        return 0;
    }

    运行结果:

    target index: 1
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  • 原文地址:https://www.cnblogs.com/pursuiting/p/10429114.html
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