【leetcode】1. Two Sum

最新推荐文章于 2025-08-13 22:03:49 发布

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最新推荐文章于 2025-08-13 22:03:49 发布

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分类专栏： leetcode 文章标签： leetcode

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本文解析了一道经典算法题“两数之和”，介绍了两种解决方案：双层循环法和利用哈希表的方法。通过对比，展示了如何将时间复杂度从O(n^2)优化至O(n)，并附带了C++实现代码。

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一、题目描述

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.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

UPDATE (2016/2/13):
The return format had been changed to zero-based indices. Please read the above updated description carefully.

题目解读：给一个整数序列和一个目标值，找出序列中两个数加起来等于目标值，返回这两个数在序列中的下标

思路：一开始只能想到写两个for循环遍历

c++代码（758ms，11.48%）就这样还能排到11%呢 = = ！汗....

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> result;
        int left,right;
        for(left=0; left<nums.size(); left++){
            for(right = left+1; right<nums.size(); right++){
                if((nums[left]+nums[right]) == target){
                    result.push_back(left);
                    result.push_back(right);
                    return result;
                }//if
            }//for
        }//for
        return result;
    }
};

看讨论区别人的代码，在遍历的同时使用map存储已遍历过的数的位置，以便于快速查找，时间复杂度可以达到o(n)

c++代码（16ms，62.06%）

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        //Key is the number and value is its index in the vector.
        	unordered_map<int, int> hash;
        	vector<int> result;
        	for (int i = 0; i < numbers.size(); i++) {
        		int numberToFind = target - numbers[i];
        
                    //if numberToFind is found in map, return them
        		if (hash.find(numberToFind) != hash.end()) {
                            //+1 because indices are NOT zero based
        			result.push_back(hash[numberToFind]);
        			result.push_back(i);			
        			return result;
        		}
        
                    //number was not found. Put it in the map.
        		hash[numbers[i]] = i;
        	}
        	return result;
    }
};