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1. Two Sum - Easy

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].

 

用hashmap記錄陣列元素及其下標 <nums[i], i>

注意返回值要求是下標小的在前面,而從左往右遍歷陣列, 一般在右半部分才能找到相加為target的數,此時的index應該是較的的那個,應該放在第二位,從map中取得的index應該放在第一位。

時間:O(N),空間:O(N)

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