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].
My solution: brute force
class Solution {
public int[] twoSum(int[] nums, int target) {
int[] result = new int[2];
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;
}
}
}
return null;
}
}
test result:
Runtime: 33 ms, faster than 28.56% of Java online submissions for Two Sum.
The below solution is provided by leetcode solutions:
Approach 2: Two-pass Hash Table
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
map.put(nums[i], i);
}
for (int i = 0; i < nums.length; i++) {
int complement = target - nums[i];
if (map.containsKey(complement) && map.get(complement) != i) {
return new int[] { i, map.get(complement) };
}
}
throw new IllegalArgumentException("No two sum solution");
}
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
int complement = target - nums[i];
if (map.containsKey(complement)) {
return new int[] { map.get(complement), i };
}
map.put(nums[i], i);
}
throw new IllegalArgumentException("No two sum solution");
}
Summary:
The first attempt to the leetcode Q.No.1 succeed with pretty low efficiency.
I just felt a bit depressed ? (kidding)
The other two solutions has amazed me by its running speed. Somehow using hashmap
is kind of cheating but is a smart solution.