LeeCode编程训练日记一：Two Sum

通过Leecode训练编程，每天只做一个题目，重在训练编程思想

Practice Day 1: 2017/06/25 

参考：http://blog.csdn.net/murdock_c/article/details/49851547

test 1:

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].

解答：

1、暴力计算(不推荐，时间复杂度太高)

#include<iostream>using namespace std;int main() {     int arr[4] = {2,7,11,15}

     int target;

     int a,b;      cin >> target;

     int length = sizeof(arr)/sizeof(arr[0]);

     for(int i = 0; i < length; i++){

for(int j = i+1; j< length; j++) {

if(arr[i] + arr[j] == target)

{

a = i;

b = j;

break;

     }

  }

     } 

    cout << arr[a] << " & " << arr[b] << endl;

    return 0; }

2、c++ map的方法来做，时间复杂度因为map查找速度快的原因而大大下降

#include<iostream>#include<vector>#include<map>using namespace std;class Solution {public:      vector<int> twosome(vector<int> numbers, int target) {      int i, sum;      vector<int> results;

      map<int, int> hmap;

      for(int i=0; i < numbers.size(); i++)

{

   if(!hmap(numbers[i])){

map.insert(pair<int,int>(numbers[i], i));

   }

           if(map.count(target - numbers[i]))

{

   int n = hmap[target - numbers[i]];

   if(n < i)

{

   results.push_back(n);

   results.push_back(i);

   return results;

}

     }

  } 

     return results;}};

int main() {

vector<int> numbers = {2,7,11,15};

vector<int> res;

int target;

cin >> target;

Solution solution;

res = solution.twosum(numbers, target);

cout << " 返回的下标的索引分别为：";
for(int i = 0; i < 2; i++)
{
cout << res[i] << "," 
}
return 0;
}

if(!hmap(numbers[i])){
map.insert(pair<int,int>(numbers[i], i));
这句意思就是如果numbers[i]在map中没有出现过，就将（numbers[i],i）插入到map中，其中numbers[i]相当于key，i相当于value。因为map中的查找是很快的，所以时间复杂度很低。 
另外，map要求其中的key值不允许重复，即一个key只对应一个value。