0001. Two Sum
Easy | Hash Table | 28 ms (93.73%), 13.4 MB (79.11%)
Source: LeetCode - Two Sum GitHub: Solution / Performance
Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
You can return the answer in any order.
Constraints:
2 <= nums.length <= 10^4-10^9 <= nums[i] <= 10^9-10^9 <= target <= 10^9Only one valid answer exists.
Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Output: Because nums[0] + nums[1] == 9, we return [0, 1].
Input: nums = [3,2,4], target = 6
Output: [1,2]
Input: nums = [3,3], target = 6
Output: [0,1]Each number in the list of integers has its own "remain" value by subtracting its value from the target number.
So that we could iterate each element in the list of integers by the index, and then check whether the value of remain exists.
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
hashTable = dict()
# ==================================================
# Array + Hash Table =
# ==================================================
# time : O(n)
# space : O(n)
for i in range(len(nums)):
remain = target - nums[i]
if remain in hashTable:
return [i, hashTable[remain]]
# We could overwrite since there is only one solution
hashTable[nums[i]] = iclass Solution {
/**
* @time : O(n)
* @space : O(n)
*/
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> map = new HashMap<>();
for(int i=0 ; i<nums.length ; i++) {
int remain = target - nums[i];
if(map.containsKey(remain)) {
return new int[] {i, map.get(remain)};
}
map.put(nums[i], i);
}
return null;
}
}Last updated
Was this helpful?