leetcode 217 Contains Duplicate 数组中是否有重复的数字

class Solution {
public:
    bool containsDuplicate(vector<int>& nums)
    {
        set<int> result;
        
         set<int>::iterator itor ;

    for(int i = 0;i< nums.size();++i)
    {
        itor = result.find(nums[i]) ;

        if(itor != result.end()) 
        {
            return true;
        }
        else
        {
            result.insert(nums[i]);
        }
    }
    
    return false;
 

        
    }
};

class Solution {
public:
    bool containsDuplicate(vector<int>& nums) {
        return nums.size() > set<int>(nums.begin(), nums.end()).size();        
    }
};

python 的版本：

class Solution:
    def containsDuplicate(self, nums):
        return len(nums) > len(set(nums))

c++ 的hash版本：同类的hash code是相同的，这是一个非常重要的编程思想

class Solution {
public:
    bool containsDuplicate(vector<int>& nums) {
        unordered_set<int> hashset;
        for (int i = 0; i < nums.size(); ++i) {
            if (hashset.find(nums[i]) != hashset.end()) {
                return true;
            }
            else {
                hashset.insert(nums[i]);
            }
        }
        return false;
    }
};

c++排序版本：

+2 votes
942 views
class Solution {
public:
    bool containsDuplicate(vector<int>& nums) 
    {
        int size=nums.size();
        sort(nums.begin(),nums.end());
        nums.erase(unique(nums.begin(),nums.end()),nums.end());
        return (size!=nums.size());
    }
};

+4 votes
Your running time is 28ms, if not use unique, it will be 24ms:
class Solution {
public:
    bool containsDuplicate(std::vector<int>& nums) {
        std::sort(nums.begin(), nums.end());
        for (int i = 1; i < nums.size(); ++i)
            if (nums[i] == nums[i - 1])
                return true;
        return false;
    }
};