Two Sum LT1

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].

Idea: map to store index

Note. 不要忘记加元素进去Map, don't forget o add element to update map.

Time complexity: O(n)

Space complexity: O(n)

 class Solution {
     public int[] twoSum(int[] nums, int target) {
         Map<Integer, Integer> record = new HashMap<>();
         for(int i = 0; i < nums.length; ++i) {
             Integer index = record.get(target - nums[i]);
             if(index != null) {
                 return new int[] {index, i};
             }
             record.put(nums[i], i);
         }

         return new int[0];
     }
 }

python

 class Solution:
     def twoSum(self, nums: List[int], target: int) -> List[int]:
         record = {}
         for i in range(len(nums)):
             if target - nums[i] in record:
                 return [record[target-nums[i]], i]
             record[nums[i]] = i
         return []