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].
Solution1: Brute-force
Lock pointer i, move pointer j to check if nums[j] == target - nums[i].
code:
1 /* Time Complexity: O(n*n) 2 Space Complexity: O(1) 3 */ 4 class Solution { 5 public int[] twoSum(int[] nums, int target) { 6 for (int i = 0; i < nums.length; i++) { 7 for (int j = i + 1; j < nums.length; j++) { 8 if (nums[j] == target - nums[i]) { 9 return new int[] { i, j }; 10 } 11 } 12 } 13 return null; 14 } 15 }
Solution2: HashMap
Use a map to record the index of each element and check if target - nums[i] is in the map. if so, we find a pair whose sum is equal to target.
code:
1 /* Time Complexity: O(n) 2 Space Complexity: O(n) 3 */ 4 class Solution { 5 public int[] twoSum(int[] nums, int target) { 6 Map<Integer, Integer> map = new HashMap<>(); 7 for(int i = 0; i< nums.length; i++){ 8 if(map.containsKey(target - nums[i])){ 9 return new int[]{i, map.get(target - nums[i])}; 10 }else{ 11 map.put(nums[i], i); 12 } 13 } 14 return null; 15 } 16 }