[LeetCode]Two Sum

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

import java.util.HashMap;
public class Solution {
static public int[] twoSum(int[] numbers, int target) {
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
for (int i=0; i<numbers.length; i++){
map.put(numbers[i], i+1);
}
for (int i=0; i<numbers.length; i++){
//!=null:has two value can achieve the target by adding
//>i+1: make sure index1 < index2
if (map.get(target - numbers[i]) != null && map.get(target - numbers[i]) > i+1){
int[] result = new int[2];
result[0] = i+1;//index of smaller one
result[1] = map.get(target - numbers[i]);//index of larger one
return result;
}
}
return null;
}

public static void main(String[] args){
int[] numbers={2, 7, 15, 11};
int target=9;
int[] n = twoSum(numbers,target);
System.out.println(n[0] + "," + n[1]);
}
}

O(n2) runtime, O(1) space – Brute force:

The brute force approach is simple. Loop through each element x and find if there is another value that equals to target – x. As finding another value requires looping through the rest of array, its runtime complexity is O(n2).

O(n) runtime, O(n) space – Hash table:

We could reduce the runtime complexity of looking up a value to O(1) using a hash map that maps a value to its index.

Other solutions:

First quicksort(array), then

public int[] twoSum(int[] numbers, int target) {
int i = 0;
int j = numbers.length-1;
while(i<j){
if(numbers[i] + numbers[j] == target){
int[] twoIndex = {i+1,j+1};
return twoIndex;
}
else if(numbers[i] + numbers[j] > target){
j--;
}
else{
i++;
}
}
return null;
}

T(n) = O(nlgn) + O(n)