Leetcode Two Sum

最新推荐文章于 2025-09-10 15:16:07 发布

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原文链接：http://www.cnblogs.com/xiongqiangcs/p/3620024.html

文章标签：

#c/c++ #java

本文介绍了一种解决两数之和问题的有效算法，通过排序和双指针技术找到目标和的两个数，并讨论了使用map和unordered_map两种数据结构进行优化的方法。

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

//先对numbers数组排序，然后利用左右两个指针查找，注意原有数组的索引，时间复杂度O(nlogn)
#include <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <iterator>
using namespace std;

vector<int> twoSum(vector<int> &numbers, int target){
    vector<int> numberCopy=numbers;
    sort(numberCopy.begin(),numberCopy.end());
    int left = 0, right = numberCopy.size()-1;
    vector<int> res;
    while(left < right){
        if(numberCopy[left]+numberCopy[right] > target) right--;
        else if(numberCopy[left]+numberCopy[right] < target) left++;
        else {
            vector<int>::iterator firstIndex = find(numbers.begin(),numbers.end(),numberCopy[left]);
            int first = distance(numbers.begin(),firstIndex);
            vector<int>::iterator secondIndex = find(numbers.begin(), numbers.end(),numberCopy[right]);
            if(secondIndex == firstIndex) secondIndex = find(secondIndex+1, numbers.end(),numberCopy[right]);
            int second = distance(numbers.begin(),secondIndex);
            if(first > second) swap(first,second);
            res.push_back(first+1);
            res.push_back(second+1);
            break;
        }
    }
    return res;
}

int main(){
    vector<int> numbers;
    numbers.push_back(0);
    numbers.push_back(4);
    numbers.push_back(3);
    numbers.push_back(0);
    vector<int> res = twoSum(numbers,0);
    cout<<res[0]<<" "<<res[1]<<endl;
}

/*
 *利用map存储，由于map的find函数的时间复杂度是O(logn)，故总体的时间复杂度是O(nlogn),
 *可以将map换成hashMap，hashMap的查找时间复杂度为O(1),故整体时间复杂度为O(n)
 *由于C++98不支持hashMap可以换成java的hashMap
 *也可以利用C++11的unordered_map,其是c++的hashMap
 */
 
#include <iostream>
#include <vector>
#include <map>

using namespace std;

vector<int> twoSum(vector<int> &numbers, int target){
    map<int,int> twoSumMap;
    vector<int> res(2);
    for(int i = 0 ; i <  numbers.size(); ++ i){
        if(twoSumMap.find(numbers[i])!= twoSumMap.end()){
            res[0] = twoSumMap[numbers[i]]+1;
            res[1] = i + 1;
            break;
        }else{
            twoSumMap.insert(make_pair(target-numbers[i],i));
        }
    }
    return res;
}

利用C++11的unordered_map,时间复杂度O(n)

#include <iostream>
#include <vector>
#include <unordered_map>

using namespace std;

vector<int> twoSum(vector<int> &numbers, int target){
    unordered_map<int,int> twoSumMap;
    vector<int> res(2);
    for(int i = 0 ; i <  numbers.size(); ++ i){
        if(twoSumMap.find(numbers[i])!= twoSumMap.end()){
            res[0] = twoSumMap[numbers[i]]+1;
            res[1] = i + 1;
            break;
        }else{
            twoSumMap.insert(make_pair(target-numbers[i],i));
        }
    }
    return res;
}