Description
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].
Solution
Approach 1: Brute Force
The easiest way is to use brute force to traverse all the elements in the array to get the solution.
public int[] twoSum(int[] nums, int target) {
for (int i = 0; i < nums.length; i++) {
for (int j = i + 1; j < nums.length; j++) {
if (nums[j] + nums[i] == target) {
return new int[] { i, j };
}
}
}
}
The brute force method can solve this problem, but its time complexity is O(n2).
Approach 2: Hash Table
A better solution is to use Hash Table to solve this problem, only need to traverse the array once, so its time complexity is O (n).
public int[] twoSum(int[] nums, int target) {
int[] ret = new int[2];
var map = new HashMap<Integer, Integer>();
int remainder;
for (int i = 0; i < nums.length; i++) {
remainder = target - nums[i];
if (map.containsKey(remainder)) {
ret[0] = map.get(remainder);
ret[1] = i;
break;
}
map.put(nums[i], i);
}
return ret;
}