[LeetCode] 120: Two Sum

最新推荐文章于 2022-01-20 20:39:30 发布

1.02^365的成长裂变

最新推荐文章于 2022-01-20 20:39:30 发布

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分类专栏： LeetCode2

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本文提供了一种解决两数之和问题的有效算法。通过使用哈希表存储数组元素及其索引，可以在O(n)的时间复杂度内找到两个数，使它们的和等于给定的目标值。文章详细介绍了算法实现过程，并附带了完整的C++代码。

[Problem]

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

[Solution]

class Solution {
public:
	vector<int> twoSum(vector<int> &numbers, int target) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function

		vector<int> res;
		// empty
		if(numbers.size() == 0)return res;

		// store into map
		map<int, vector<int> > myMap;
		for(int i = 0; i < numbers.size(); ++i){
			if(myMap.find(numbers[i]) == myMap.end()){
				vector<int> tmp(1, i+1);
				myMap[numbers[i]] = tmp;
			}
			else{
				myMap[numbers[i]].push_back(i+1);
			}
		}

		// get result
		sort(numbers.begin(), numbers.end());
		int i = 0, j = numbers.size()-1;
		while(i < j){
			if(numbers[i] + numbers[j] == target){
				if(numbers[i] != numbers[j]){
					if(myMap[numbers[i]][0] < myMap[numbers[j]][0]){
						res.push_back(myMap[numbers[i]][0]);
						res.push_back(myMap[numbers[j]][0]);
					}
					else{
						res.push_back(myMap[numbers[j]][0]);
						res.push_back(myMap[numbers[i]][0]);
					}
				}
				else{
					res.push_back(myMap[numbers[i]][0]);
					res.push_back(myMap[numbers[j]][1]);
				}
				break;
			}
			else if(numbers[i] + numbers[j] < target){
				i++;
			}
			else{
				j--;
			}
		}
		return res;
	}
};