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记录数组元素及其下标 <nums[i], i>
注意返回值要求是下标小的在前面,而从左往右遍历数组, 一般在右半部分才能找到相加为target的数,此时的index应该是较的的那个,应该放在第二位,从map中取得的index应该放在第一位。
时间:O(N),空间: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; } }