LeetCode Two Sum

最新推荐文章于 2024-03-02 21:28:00 发布

kassical

最新推荐文章于 2024-03-02 21:28:00 发布

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分类专栏： LeetCode 文章标签： vector leetcode 数据结构

本文链接：https://blog.youkuaiyun.com/kassical/article/details/37872847

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本文介绍了一种解决两数之和问题的有效算法。给定一个整数数组及目标值，该算法能找出两个数相加等于目标值的索引。通过定义结构体保存数值及其原始索引并进行排序，利用首尾指针逼近的方法找到答案，避免了传统O(N^2)的时间复杂度。

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

注：时间复杂度为O（N*N)的通过不了，所以采用排序，然后首尾逼近，时间复杂度就是排序的O(N*logN），因为排序后数据标号变了，所以定义了一个结构体保存数据及其下标。

struct iNode {
	int num;
	int index;
};

bool cmp(const iNode &a, const iNode &b)
{
	return a.num < b.num;
}

class Solution {
public:
	vector<int> twoSum(vector<int> &numbers, int target) {
		vector<int> result;
		vector<iNode> ivec;
		for (int i = 0; i < numbers.size(); i++) {
			iNode node;
			node.num = numbers[i];
			node.index = i;
			ivec.push_back(node);
		}

		sort(ivec.begin(), ivec.end(), cmp);

		int i = 0, j = ivec.size() - 1, sum = 0;
		while (i != j) {
			sum = ivec[i].num + ivec[j].num;
			if (sum == target) {
				if (ivec[i].index < ivec[j].index) {
					result.push_back(ivec[i].index + 1);
					result.push_back(ivec[j].index + 1);
				} 
				else {
					result.push_back(ivec[j].index + 1);
					result.push_back(ivec[i].index + 1);
				}
				break;
			}
			else if (sum > target)
				j--;
			else
				i++;
		}
		return result;
	}
};