1. Two Sum

Posted 2020-07-19 早杰

Total Accepted: 238787 Total Submissions: 999694 Difficulty: Easy

 

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.

Example:

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

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

 

UPDATE (2016/2/13):
The return format had been changed to zero-based indices. Please read the above updated description carefully.

 

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 1 class Solution {
 2 public:
 3     vector<int> twoSum(vector<int>& nums, int target) {
 4         vector<int> v = nums;
 5         vector<int> w;
 6         sort(nums.begin(), nums.end());
 7         int i = 0, j = nums.size() - 1;
 8         while (i < j){
 9             if (nums[i] + nums[j] == target){
10                 for (int m = 0; m < v.size(); m++){
11                     if (v[m] == nums[i]){
12                         w.push_back(m);
13                         continue;
14                     }
15                     if (v[m] == nums[j]){
16                         w.push_back(m);
17                     }
18                 }
19                 sort(w.begin(), w.end());
20                 break;
21             }
22             else if (nums[i] + nums[j] < target){
23                 i++;
24             }
25             else{
26                 j--;
27             }
28         }
29         return w;
30     }
31 };

 1 class Solution {
 2 public:
 3     vector<int> twoSum(vector<int>& nums, int target) {
 4         vector<int> v;
 5         map<int, int> m;
 6         map<int, int>::iterator iter;
 7         for (int i = 0; i < nums.size(); i++){
 8             int complement = target - nums[i];
 9             iter = m.find(complement);
10             if(iter != m.end())
11             {
12                 v.push_back(m[complement]);
13                 v.push_back(i);
14                 break;
15             }
16             m[nums[i]] = i;
17         }
18         return v;
19     }
20 };