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.
Example:
Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].
Solution 1:
Approach: Brute force
Time complexity: O(n^2)
Space complexity: O(1)
Approach: Brute force
Time complexity: O(n^2)
Space complexity: O(1)
public class Solution { public int[] twoSum(int[] nums, int target) { int[] result=new int[2]; if(nums==null||nums.length==0) return result; for(int i=0; i<nums.length; i++){ for(int j=i+1; j<nums.length; j++){ if(nums[i]+nums[j]==target){ result[0]=i; result[1]=j; } } } return result; } }
Approach: Hash Table
Time Complexity: O(n)
Space Complexity: O(n)
public class Solution { public int[] twoSum(int[] nums, int target) { int[] result=new int[2]; if(nums==null||nums.length==0) return result; HashMap<Integer, Integer> hashMap=new HashMap<Integer, Integer>(); for(int i=0; i<nums.length; i++){ if(hashMap.containsKey(target-nums[i])){ result[0]=i; result[1]=hashMap.get(target-nums[i]); }else{ hashMap.put(nums[i], i); } } return result; } }
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