leetcode第一题(medium)——Two Sum-优快云博客

本文链接：https://blog.youkuaiyun.com/zenggouzai/article/details/45872041

本文介绍了一个算法，用于在给定数组中找到两个数，它们相加等于特定的目标数。提供了两种方法，一种时间复杂度为O(n^2)，另一种优化后的算法时间复杂度为O(n)，通过使用哈希表实现。详细解释了每种方法的原理、步骤，并附带了代码示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目描述如下：

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

看到题目第一眼。首先想到的是进行两次循环来找出两个数之和来满足target，当然这是最自然的一种想法，时间复杂度为O(n^2)

public static int[] twoSum(int[] nums, int target) {
	int[] ret = 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) {
				ret[0] = i + 1;
				ret[1] = j + 1;
			}
		}
	}
	return ret;
}

第二种相对于第一种解法简便很多，时间复杂度为O(n)：

public int[] twoSum(int[] nums, int target) {
	HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
	int[] result = new int[2];
 
	for (int i = 0; i < nums.length; i++) {
		if (map.containsKey(nums[i])) {
			int index = map.get(nums[i]);
			result[0] = index+1 ;
			result[1] = i+1;
			break;
		} else {
			map.put(target - nums[i], i);
		}
	}
 
	return result;
    }