leetcode1. Two Sum

原创于 2017-09-05 22:28:32 发布 · 221 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

leetcode 专栏收录该内容

187 篇文章

订阅专栏

本文介绍了TwoSum问题的两种解决方法：暴力法和哈希表法。暴力法的时间复杂度为O(n^2)，空间复杂度为O(1)；哈希表法则将时间复杂度优化至O(n)，但空间复杂度提升到O(n)。通过具体示例说明了如何找到数组中两个数相加等于特定目标值的下标。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1. Two Sum

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].

解法一

暴力

class Solution {
    /**
     * @time O(n^2)
     * @space O(1)
     */
    public int[] twoSum(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return null;
        }

        for (int i = 0; i < nums.length; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                if (nums[i] + nums[j] == target) {
                    return new int[] {i, j};
                }
            }
        }

        return null;
    }
}

解法二

hashtable

public class Solution {
    /*
     * @time O(n)
     * @space O(n)
     */
    public int[] twoSum(int[] nums, int target) {
        Map<Integer, Integer> hashMap = new HashMap<>();

        for (int i = 0; i < nums.length; i++) {
            if (hashMap.containsKey(nums[i])) {
                int[] result = {hashMap.get(nums[i]), i};
                return result;
            } 
            hashMap.put(target - nums[i], i);
        }

        int[] result = {};
        return result;
    }
}