Leetcode | | The Sum of

In solving this problem, the first thought was to use two pointers, one in front, and the other followed. The trailing pointer then moves backwards until a matching element is found or the end of the array is reached. Continue with the next iteration. The code is as follows:

#include <iostream>#include <vector>using namespace STD;classSolution { Public: vector<int>Twosum ( vector<int>& Nums,intTarget) { vector<int>:: Iterator Iter1, iter2; vector<int>Result        Iter1 = Nums.begin (); Iter2 = Nums.begin (); for(; Iter1! = Nums.end ()-1;            iter1++) {result.clear (); Result.push_back ((Iter1-nums.begin ()) +1); for(Iter2 = Iter1 +1; Iter2! = Nums.end (); iter2++) {if(*iter1 + *iter2 = = target) {Result.push_back (Iter2-nums.begin ()) +1);returnResult }            }        }    }};intMain () { vector<int>Nums Nums.push_back (3); Nums.push_back (2); Nums.push_back (4); Nums.push_back ( the); Solution solution;cout<< Solution.twosum (Nums,7)[0] << Endl;cout<< Solution.twosum (Nums,7)[1] << Endl;  GetChar (); }

The result after submission is time limit exceed. Because the time efficiency of the algorithm is θ( n 2 ) , so the running time is too long.

The basic operation we do in the above code is to continuously compare whether the subsequent element is equal to a value, and convert the comparison operation to a lookup operation, that is, to find the value of the element that follows. For the map container, the lookup operation can be θ ( l g n ) Or θ(1) , so this will improve the time efficiency of the algorithm.

Useful Points of knowledge: 1. Convert the comparison operation into a find operation. 2. If you use the Map type container in C + + because she is using a tree at the bottom, the time efficiency of its lookup operation is θ ( l g n ) If you use the Unordered_map type container in C + + because she is using a hash table at the bottom, the time efficiency of its lookup operation is θ(1) ；

The code for using the map type container is as follows, with a run time of 24ms:

classSolution { Public: vector<int>Twosum ( vector<int>& Nums,intTarget) { map<int, int>Tree_map; vector<int>ResultintLength = Nums.size (); for(inti =0; i < length; i++) {if(Tree_map.count (Target-nums[i])) {Result.push_back (tree_map[target-nums[i]) +1); Result.push_back (i +1);returnResult        } Tree_map[nums[i]] = i; }    }};

The code for using the Unordered_map type container is as follows, with a run time of 16ms:

classSolution { Public: vector<int>Twosum ( vector<int>& Nums,intTarget) { unordered_map<int, int>Hash_map; vector<int>ResultintLength = Nums.size (); for(inti =0; i < length; i++) {if(Hash_map.count (Target-nums[i])) {Result.push_back (hash_map[target-nums[i]) +1); Result.push_back (i +1);returnResult        } Hash_map[nums[i]] = i; }    }};

Here is a reference article:
Refer to Article One
Reference Article Two

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Leetcode | | The Sum of