LeetCode - 1.Two Sum
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 (C++)
/**
* Brute-Force
* Time Complexity: O(n2)
*/
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
vector<int> result;
for (int i = 0; i < nums.size(); i++) {
for(int j = i + 1; j < nums.size(); j++) {
if(nums[i] + nums[j] == target) {
result.push_back(i);
result.push_back(j);
return result;
}
}
}
return result;
}
};
Solution 2 (C++)
/**
* Use a map to store visited numbers. For each number in nums, find target - nums[i] in map.
* Time Complexity: O(n)
*/
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
unordered_map<int, int> map;
for (int i = 0; i < nums.size(); i++) {
auto it = map.find(target - nums[i]);
if (it == map.end()) {
map.emplace(nums[i], i);
} else {
return vector<int> { it->second, i };
}
}
return vector<int> {};
}
};