题目描述
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.
样例
Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].
思路分析
一开始我只想到了两次for循环直接求解,复杂度是O(n2),之后看了讨论区,有巧用hsahmap的,可以达到O(n)。另外吐槽下,200k的答案居然有错,不能ac,委实有点尴尬好吧。
代码
方法一
public int[] twoSum(int[] nums, int target) { int []a=new int [2]; for(int i=0;i<nums.length;i++) { for(int j=i+1;j<nums.length;j++) { if(nums[i]+nums[j]==target) { a[0]=i; a[1]=j; break; } } } return a;
方法二:巧用hashmap
public int[] twoSum(int[] numbers, int target) { int[] result = new int[2]; Map<Integer, Integer> map = new HashMap<Integer, Integer>(); for (int i = 0; i < numbers.length; i++) { if (map.containsKey(target - numbers[i])) { result[1] = i ; result[0] = map.get(target - numbers[i]); return result; } map.put(numbers[i], i ); } return result; }