Leetcode 1 Two Sum

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

给定一个整数数组和一个目标值，找出数组中和为目标值的两个数。

你可以假设每个输入只对应一种答案，且同样的元素不能被重复利用。

示例:

给定 nums = [2, 7, 11, 15], target = 9

因为 nums[0] + nums[1] = 2 + 7 = 9
所以返回 [0, 1]

C++ 穷举法 (3ms)

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> result; 
        for(int i=0; i<nums.size(); i++)
            for (int j=i+1; j<nums.size();j++){
                if(nums[i]+nums[j]==target)
                {
                    result.push_back(i);
                    result.push_back(j);
                    break;
                }
            }
    return result;
    }
};

C++ 缓存映射 (2ms)

#include <vector>
#include <map>
#include <iostream>
class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
       vector<int> results;
       map<int,int> cache;
       bool find=false;
       
       for(unsigned int i=0;i<nums.size(); ++i)
       {
           cache[nums[i]]=i;
       }
        
       for(unsigned int i=0; i<nums.size(); ++i)
       {
           map<int,int>::const_iterator it=cache.find(target-nums[i]);
           if( it!= cache.end()){
               if (i != it->second)
               {
                       results.push_back(i);
                       results.push_back(it->second);
                       find = true;
                       break;
               }
               if(find)
               {
                   break;
               }
           }
       }
    return results;
    }
};

Python 穷举法 (49ms)

class Solution(object):
    def twoSum(self, nums, target):
    
        index=[]
        for i in range(0,len(nums)):
            for j in range(i+1,len(nums)):
                if (nums[i]+nums[j]==target):
                    index.append(i)
                    index.append(j)
                    break
        return index

Python dictionary （42ms）

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        
        index=[]
        dict1={}
        for i in range(len(nums)):
            dict1[nums[i]]=i
        
        for k in range(len(nums)):
            it=dict1.get(target-nums[k])
            if (k != it):
                index.append(k)
                index.append(it)
                break
            else:
                k=k+1
        return index