LeetCode 15. 三数之和 (时间超出限制后,参考他人答案)

本文探讨了LeetCode上经典问题“三数之和”的多种解法,包括一种效率低下的初试解法及一种高效的排序双指针解法。通过对代码的分析,对比了不同方法的时间复杂度,强调了排序和双指针技巧在解决此类问题中的关键作用。

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15. 三数之和

 

 

 

给定一个包含 n 个整数的数组 nums,判断 nums 中是否存在三个元素 a,b,c ,使得 a + b + c = 0 ?找出所有满足条件且不重复的三元组。

注意:答案中不可以包含重复的三元组。

例如, 给定数组 nums = [-1, 0, 1, 2, -1, -4],

满足要求的三元组集合为:
[
  [-1, 0, 1],
  [-1, -1, 2]
]

您是否在真实的面试环节中遇到过这道题目呢?

 

 

错误答案1(耗时太长):

class Solution {

public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<int> tmp;
        vector<vector<int>> res;             
        int len = nums.size();
        int t,t1,t2,t3;
        
        if(3>nums.size()){return res;}
        
        
        for(int i=0;i<len;i++)
        {
            for(int j=i+1;j<len;j++)
            {
                for(int k=j+1;k<len;k++)
                {
                    if(0==nums[i]+nums[j]+nums[k])
                    {
                        t1=nums[i];
                        t2=nums[j];
                        t3=nums[k];
                        
                        if(t1>t2)
                        {
                            t=t2;
                            t2=t1;
                            t1=t;
                        }
                        if(t2>t3)
                        {
                            t=t3;
                            t3=t2;
                            t2=t;
                        }
                        if(t1>t2)
                        {
                            t=t2;
                            t2=t1;
                            t1=t;
                        }
                        tmp.clear();
                        tmp.push_back(t1);
                        tmp.push_back(t2);
                        tmp.push_back(t3);
                        
                        if(0!=res.size())
                        {
                            t=0;
                            for(int n=0;n<res.size();n++)
                            {
                                if(tmp == res[n])
                                {
                                    t = 1;
                                    break;
                                }
                            }
                            if(!t)
                            {
                                res.push_back(tmp);
                            }
                        }
                        else
                        {
                            res.push_back(tmp);
                        }
                    }
                }
            }   
        }
        
        return res;
    }

};

 

 

提交记录

311 / 313 个通过测试用例

状态:

超出时间限制

 

提交时间:0 分钟之前

最后执行的输入: [82597,-9243,62390,83030,-97960,-26521,-61011,83390,-38677,12333,75987,46091,83794,19355,-71037,-6242,-28801,324......

 

 

 

我的答案2:

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<int> tmp;
        vector<vector<int>> res;             
        int len = nums.size();
        int t,t1,t2,t3;
        
        
        vector<int>::iterator iter_i;
        vector<int>::iterator iter_j;
        vector<int>::iterator iter_k;
        vector<vector<int>>::iterator iter_res;
        
        iter_i = nums.begin();  
        for(int i=0;i<len;i++)
        {
            iter_j = nums.begin();
            iter_j += i+1;
            for(int j=i+1;j<len;j++)
            {
                iter_k = nums.begin();
                iter_k += j+1;
                for(int k=j+1;k<len;k++)
                {
                    if(0==*iter_i+*iter_j+*iter_k)
                    {
                        t1=*iter_i;
                        t2=*iter_j;
                        t3=*iter_k;
                        
                        if(t1>t2)
                        {
                            t=t2;
                            t2=t1;
                            t1=t;
                        }
                        if(t2>t3)
                        {
                            t=t3;
                            t3=t2;
                            t2=t;
                        }
                        if(t1>t2)
                        {
                            t=t2;
                            t2=t1;
                            t1=t;
                        }
                        tmp.clear();
                        tmp.push_back(t1);
                        tmp.push_back(t2);
                        tmp.push_back(t3);
                        
                        if(0!=res.size())
                        {
                            t=0;
                            
                            for(iter_res = res.begin();iter_res!=res.end();iter_res++)
                            {
                                if(tmp == *iter_res)
                                {
                                    t = 1;
                                    break;
                                }
                            }
                            if(!t)
                            {
                                res.push_back(tmp);
                            }
                        }
                        else
                        {
                            res.push_back(tmp);
                        }
                    }
                    iter_k++;
                }
                iter_j++;
            }
            iter_i++;
        }
        
        return res;
    }
};

 

依旧是超出时间限制!!

 

于是乎在网上找了参考答案:

 

来源:https://blog.youkuaiyun.com/shinanhualiu/article/details/48913435

思路:solution:
因为三元组的元素是按照递增的顺序,因此我们首先需要对S进行排序。然后保持两个前后索引来进行比较。

class Solution {
public:    
    /**
     * @param numbers : Give an array numbers of n integer
     * @return : Find all unique triplets in the array which gives the sum of zero.
     */
    vector<vector<int> > threeSum(vector<int> &nums) {
        // write your code here   
        vector<vector<int>> res;
        sort(nums.begin(), nums.end()); //对数组进行排序
        int i = 0 , last = nums.size()-1;
        while (i < last) {
            int a = nums[i], j = i + 1, k = last;
            while (j < k) {
                int b = nums[j], c = nums[k];
                int sum = a + b + c;
                if (sum == 0)  res.push_back({a,b,c});
                if (sum <= 0)  //注意不能含有相同的索引
                    while (nums[j] == b && j < k ) ++j;
                if (sum >= 0)
                    while (nums[k] == c && j < k) --k;
            }
            while (nums[i] == a && i < last) ++i;
        }
        return res;        
    }
};

 

果然没有好好思考出最好的方式,做了很多已然一看就和不会为0的遍历,真是笨 = = ;

 

 

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