Introduction
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].
Solution 1: hash map
Using hash map for the solution, record the needed value and the index pair.
For Example: [2,7,11,15], target = 9
Then the algorithm goes like:
- map->search 2: not find -> insert <7,0> (9-2 = 7, index 0)
- map->search 7: find -> return {map[7],1} -> {0,1}
Space Complexity: O(n)
Time Complexity: O(n)
Code:
vector<int> twoSum(vector<int>& nums, int target) {
unordered_map<int,int> dict;
for(int i = 0; i < nums.size(); ++i)
{
if(dict.find(nums[i])==dict.end())
{
dict[target-nums[i]] = i;
}else{
return {dict[nums[i]],i};
}
}
return {};
}P.S:
Two pointer is not suitable for this problem, because after sorting the array, the index is completely disorganized.
