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 []