Overview
Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target. You may assume that each input would have exactly one solution, and you may not use the same element twice. You can return the answer in any order.
Samples
Example-1
Input: nums = [2,7,11,15], target = 9 Output: [0,1] Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].
Example 2:
Input: nums = [3,2,4], target = 6 Output: [1,2]
Example 3:
Input: nums = [3,3], target = 6 Output: [0,1]
Solution-1:
public static int[] CalCulateTwoSum(int[] nums, int target) { int[] res = new int[2]; for (int i = 0; i <= nums.Length - 1; i++) { for (int j = i + 1; j <= nums.Length - 1; j++) { if (target == (nums[i] + nums[j])) { res[0] = i; res[1] = j; break; } } } return res; }Solution-2:
public static int[] CalCulateTwoSumV1(int[] nums, int target) { Dictionary<int, int> map = new Dictionary<int, int>(); for (int i = 0; i < nums.Length; i++) { map[nums[i]] = i; }
for (int i = 0; i < nums.Length; i++) { int complement = target - nums[i]; if (map.ContainsKey(complement) && map[complement] != i) { return new int[] { i, map[complement] }; } }
return null; }Solution-2:
public static int[] CalCulateTwoSumV2(int[] nums, int target) { Dictionary<int, int> map = new Dictionary<int, int>(); for (int i = 0; i < nums.Length; i++) { int complement = target - nums[i]; if (map.ContainsKey(complement)) { return new int[] { map[complement], i }; }
map[nums[i]] = i; }
return null; }Advanced Slot Usage
we are cureently working on the benchmark to provide best solution.
