LeetCode 1.Two Sum

最新推荐文章于 2020-10-21 10:33:51 发布

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本文探讨了在整数数组中寻找两个数相加等于特定目标值的问题，并提供了两种解决方案：暴力破解法与哈希表法。通过这两种方法，文章详细介绍了如何高效地找到符合条件的两个数的下标。

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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, and you may not use the same element twice.

Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

brute force

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        int i,j;
        vector<int> res;
        for(i=0;i<nums.size();i++)
            for(j=i+1;j<nums.size();j++)
                if(nums[i] + nums[j] == target)
                {
                    res.push_back(i);
                    res.push_back(j);
                    break;
                }
        return res;
    }
};

We reduce the look up time from O(n) to O(1) by trading space for speed. A hash table is built exactly for this purpose, it supports fast look up in near constant time.

hash table

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        int i,j;
        vector<int> res;
        map<int,int> notme;
        for(i=0;i<nums.size();i++)
            notme[nums[i]]=i;
        for(i=0;i<nums.size();i++)
        {
            j=target - nums[i];
            if(notme.count(j)&&j!=i)
            {
                res.push_back(i);
                res.push_back(j);
                break;
            }
        }
        return res;
    }
};

或者可以把两个for循环写一起代码简洁性增强时间复杂度不变

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        int i,j;
        vector<int> res;
        map<int,int> notme;
        for(i=0;i<nums.size();i++)
        {
            notme[nums[i]]=i;
            j = target - nums[i];
            if(notme.count(j)&&notme[j]!=i)
            {
                res.push_back(notme[j]);
                res.push_back(i);
                    break;
            }
        }

        return res;
    }
};