Leetcode Two Sum

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;
}