[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].

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。

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。

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");
}
}