[leetcode]1. Two Sum

题目链接

问题描述

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.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,

return [0, 1].

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概要

给定一个整型数组和一个目标数，返回数组中两数相加能相等于目标数的两个索引，即下标。
假设每一个输入的数组仅有一个正确结果，且每个元素不能使用两次，即不能与自身相加。

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思路

最容易的就是暴力法，逐个逐个地进行相加，只要和等于目标数，就得到结果，但比较慢。
第二种就利用哈希表来辅助查找，降低了时间复杂度，但增加了空间复杂度。

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暴力法代码实现 时间复杂度 O(n^2) 空间复杂度 O(1)

public class Solution {    public int[] twoSum(int[] nums, int target) {        int length = nums.length;        for (int i = 0; i < length; i++) {            for (int j = i + 1; j < length; j++) {                if (nums[i] + nums[j] == target) {                    return new int[]{i, j};                }            }        }        throw new IllegalArgumentException("No two sum solution");    }}

哈希表辅助法实现 时间复杂度 O(n) 空间复杂度 O(n)

public class Solution {    public int[] twoSum(int[] nums, int target) {        int length = nums.length;        Map<Integer, Integer> map = new HashMap<Integer, Integer>();        for (int i = 0; i < length; i++) {            map.put(nums[i], i);        }        for (int i = 0; i < length; i++) {            int difference = target - nums[i];            if (map.containsKey(difference) && map.get(difference) != i) {                return new int[]{i, map.get(difference)};            }        }        throw new IllegalArgumentException("No two sum solution");    }}