Leetcode116: Two Sum

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最新推荐文章于 2022-05-19 22:23:19 发布

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分类专栏： leetcode c++ 文章标签： leetcode algorithm

本文链接：https://blog.youkuaiyun.com/u013089961/article/details/49384417

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本文介绍了一种解决两数之和问题的有效算法，通过使用哈希表(unordered_map)来快速查找目标和所需的配对数字。该算法适用于整数数组中寻找两个数使它们的和等于特定的目标值，提供了C++与Python两种语言的实现方案。

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

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> res(2);
        int n = nums.size();
        map<int,int> mp;
        for(int i = 0; i < n; i++)
        {
            if(mp.find(target-nums[i]) == mp.end())
                mp[nums[i]] = i;
            else
            {
                res[0] = mp[target-nums[i]]+1;
                res[1] = i+1;
            }
        }
        return res;
    }
};

用unorderedmap比map快，一个是hash散列实现，一个是红黑树。

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
    	vector<int> res(2);
        unordered_map<int, int> mp;
        for (int i = 0; i < nums.size(); ++i)
        {
        	/* code */
        	if(mp.find(target-nums[i]) == mp.end())
        		mp[nums[i]] = i;
        	else{
        		res[0] = mp[target-nums[i]];
        		res[1] = i;
        		break;
        	}
        }
        return res;
    }
};

python代码：

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        mp={}
        for x in range(len(nums)):
        	if target-nums[x] in mp:
        		return [mp[target-nums[x]], x]
        	mp[nums[x]] = x