[LeetCode1] Two Sum

最新推荐文章于 2025-06-17 11:45:49 发布

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最新推荐文章于 2025-06-17 11:45:49 发布

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分类专栏： Leetcode LeetCode Questions 文章标签： leetcode hashmap

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本文介绍了一种经典的算法问题——寻找数组中两个数相加等于特定目标值的方法。提供了两种解决方案：一种利用哈希表实现高效查找，时间复杂度为O(n)；另一种采用暴力搜索的方式解决该问题。

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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

Analysis

1. Hash

从左往右扫描一遍，然后将数及坐标，存到map中。然后再扫描一遍即可。时间复杂度O(n)

2. 暴力搜索
这个比较省事，时间复杂度O(n*n)

java

public int[] twoSum(int[] numbers, int target) {
        int index[] = {0,0};
		HashMap<Integer, Integer> num = new HashMap<Integer,Integer>();
		for(int i=0;i<numbers.length;i++){
			num.put(numbers[i], i);
		}
		for(int i=0;i<numbers.length;i++){
			int temp = target-numbers[i];
			if(num.containsKey(temp)&&i<num.get(temp)){
				index[0] = i+1;
				index[1] = num.get(temp)+1;
				break;
			}
		}
		return index;
    }

c++

vector<int> twoSum(vector<int> &numbers, int target) {
     map<int,int> mapping;
    vector<int> result;
    for(int i=0;i<numbers.size();i++){
        mapping[numbers[i]] = i;
    }
    for(int i=0;i<numbers.size();i++){
        int searched = target-numbers[i];
        if(mapping.find(searched)!=mapping.end()){//find the element
            if(i<mapping[searched]){
                result.push_back(i+1);
                result.push_back(mapping[searched]+1);
                return result;
            }
            if(i>mapping[searched]){
                result.push_back(mapping[searched]+1);
                result.push_back(i+1);
                return result;
            }
        }
    }
    return result;
    }

solution2

java

public int[] twoSum(int[] numbers, int target) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int index[] = {0,0};
		for(int i=0; i<numbers.length; i ++){
			for(int j=i+1; j<numbers.length; j++){
				if(target == numbers[i]+numbers[j]){
					index[0] = i+1;
					index[1] = j+1;
					return index;
				}
			}
		}
		return index;
    }