Day one sum of two numbers

Day one

Sum of two numbers

Given an array of integers numsand a target value target, and you figure out a target value of the two integers in the array, and return to their array subscript.

You can assume that each input will only correspond to one answer. However, the same element in the array cannot be used twice .

Example:

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++

Here is a high praise problem solution, and its mistakes have been corrected

Violence law:

Because the same element in the array can not be used twice, so there is no similar [2,2,3,4],target=6circumstances

But do not rule out [2,5,3,4]，target=7the case, then the default is looking to return to the first answer

After thinking about the two problems above, then start thinking about solutions

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

Summary: The method is simple but the time complexity is O(n ² ). Space complexity is O(1),
slow running speed and large memory space consumption

Two-pass hash table:

class Solution {
    
    
public:
    vector<int> twoSum(vector<int>& nums, int target) {
    
    
        map<int,int> a;//建立hash表存放数组元素
        vector<int> b(2,-1);//存放结果
        for(int i=0;i<nums.size();i++)
            a.insert(map<int,int>::value_type(nums[i],i));
        for(int i=0;i<nums.size();i++)
        {
    
    
            if(a.count(target-nums[i])>0&&(a[target-nums[i]]!=i))
            //判断是否找到目标元素且目标元素不能是本身
            {
    
    
                b[0]=i;
                b[1]=a[target-nums[i]];
                break;
            }
        }
        return b;
    };
};

A hash table

Improvement: While iterating and inserting elements into the table, we will also go back and check whether the target element corresponding to the current element already exists in the table. If it exists, then we have found the corresponding solution and return it immediately.

class Solution {
    
    
public:
    vector<int> twoSum(vector<int>& nums, int target) {
    
    
        map<int,int> a;//提供一对一的hash
        vector<int> b(2,-1);//用来承载结果，初始化一个大小为2，值为-1的容器b
        for(int i=0;i<nums.size();i++)
        {
    
    
            if(a.count(target-nums[i])>0)
            {
    
    
                b[0]=a[target-nums[i]];
                b[1]=i;
                break;
            }
            a[nums[i]]=i;//反过来放入map中，用来获取结果下标
        }
        return b;
    };
};