【两次过】Lintcode 56:Two Sum-优快云博客

本文链接：https://blog.youkuaiyun.com/majichen95/article/details/79216053

本文探讨了在整数数组中寻找两个数使它们的和等于特定目标值的问题。介绍了三种解题方法：双层循环遍历、使用哈希表记录数值与下标关系以及改进版哈希表策略。每种方法都详细解释了实现思路与关键代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are zero-based.

Notice

You may assume that each input would have exactly one solution

Example

numbers=[2, 7, 11, 15], target=9

return [0, 1]

解题思路1：

最简单的就是两层循环遍历即可，没啥技术含量。

class Solution {
public:
    /*
     * @param numbers: An array of Integer
     * @param target: target = numbers[index1] + numbers[index2]
     * @return: [index1 + 1, index2 + 1] (index1 < index2)
     */
    vector<int> twoSum(vector<int> &numbers, int target)
    {
        // write your code here
        vector<int> result;
        for(int i=0;i<numbers.size();i++)
        {
            for(int j=i+1;j<numbers.size();j++)
            {
                if(numbers[i]+numbers[j] == target)
                {
                    result.push_back(i);
                    result.push_back(j);
                    return result;
                }
            }
        }
   
    }
};

解题思路2：

因为最后需要返回下标，所以将数字与下标的对应关系先存进哈希表中。防止存进哈希表中丢失下标数据。注意如果有相同数字放进哈希表时，最终只会存在最后添加的数字。

class Solution {
public:
    /**
     * @param numbers: An array of Integer
     * @param target: target = numbers[index1] + numbers[index2]
     * @return: [index1, index2] (index1 < index2)
     */
    vector<int> twoSum(vector<int> &numbers, int target) 
    {
        // write your code here
        unordered_map<int,int> record;
        for(int i = 0 ; i < numbers.size() ; i ++)
            record[numbers[i]] = i;
        
        vector<int> res;
        for(int i = 0 ; i < numbers.size() ; i ++)
        {
            int tmp = target - numbers[i];
            
            if(record.count(tmp) && record[tmp] != i)
            {
                res.push_back(i);
                res.push_back(record[tmp]);
                
                return res;
            }
        }
    }
};

解题思路3：

为了规避相同数字放入哈希表表会覆盖掉前一个数字的下标问题，这里不采用一次性放入哈希表，而采用检查一次放一次的策略。

注意下标的顺序。

class Solution {
public:
    /**
     * @param numbers: An array of Integer
     * @param target: target = numbers[index1] + numbers[index2]
     * @return: [index1, index2] (index1 < index2)
     */
    vector<int> twoSum(vector<int> &numbers, int target) 
    {
        // write your code here
        unordered_map<int,int> record;
        
        vector<int> res;
        
        for(int i = 0 ; i < numbers.size() ; i ++)
        {
            int tmp = target - numbers[i];
            
            if(record.count(tmp))
            {
                if(i < record[tmp])
                {
                    res.push_back(i);
                    res.push_back(record[tmp]);
                }
                else
                {
                    res.push_back(record[tmp]);
                    res.push_back(i);
                }
                
                return res;
            }
            
            record[numbers[i]] = i;
        }
    }
};