1. Two Sum

1. Two Sum (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, 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].

Related Topics:
Array, Hash Table

Similar Questions:

• 3Sum (Medium)
• 4Sum (Medium)
• Two Sum II - Input array is sorted (Easy)
• Two Sum III - Data structure design (Easy)
• Subarray Sum Equals K (Medium)
• Two Sum IV - Input is a BST (Easy)
• Two Sum Less Than K (Easy)

Solution 1.

// OJ: https://leetcode.com/problems/two-sum/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    vector<int> twoSum(vector<int>& A, int target) {
        vector<vector<int>> v; 
        int N = A.size(), L = 0, R = N - 1;
        for (int i = 0; i < N; ++i) v.push_back({ A[i], i });
        sort(begin(v), end(v), [](const vector<int> &a, const vector<int> &b) { return a[0] < b[0]; });
        while (L < R) {
            int sum = v[L][0] + v[R][0];
            if (sum == target) return { v[L][1], v[R][1] };
            if (sum < target) ++L;
            else --R;
        }
        return {};
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/two-sum/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    vector<int> twoSum(vector<int>& A, int target) {
        unordered_map<int, int> m;
        for (int i = 0; i < A.size(); ++i) {
            int t = target - A[i];
            if (m.count(t)) return { m[t], i };
            m[A[i]] = i;
        }
        return {};
    }
};