LeetCode

LeetCode - Two Sum

2013.12.1 02:30

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

Solution:

　　For an array of integers and a target value, find out the pair of numbers in the array that add up to the target.

　　My first solution is simply brute-force, for each element a[i], check if there exists a[j] in a[i + 1 : n] such that a[i] + a[j] == target.

　　Obviously, the time complexity is O(n^2) with double for-loop, while space complexity is O(1).

Accepted code:

class Solution {
public:
    vector<int> twoSum(vector<int> &numbers, int target) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        result.clear();
        
        int i, j, n;
        
        n = numbers.size();
        for(i = 0; i < n; ++i){
            for(j = i + 1; j < n; ++j){
                if(numbers[i] + numbers[j] == target){
                    result.push_back(i + 1);
                    result.push_back(j + 1);
                    break;
                }
            }
            if(j < n){
                break;
            }
        }
        
        return result;
    }
private:
    vector<int> result;
};

　　So the array is not sorted, why not sort it and use binary search for a faster solution? Let's do this with index sorting algorithm.

　　If the data is sorted, we can easily use binary search to find the solution with O(n * log(n)) time, but unable to guarantee that index1 < index2.

　　Thus we preserve the original data and use another array to record the indices, and sort this index array. Index sorting requires a sorting comparator, called by sort() function. I used global data for the sake of convenience, otherwise an efficient sorting algorithm must be written inside the class. Time complexity is O(n * log(n)), both in sorting and searching. Space compexity is O(n), with the $index array, and the global $data array.

Accepted code:

#include <algorithm>
using namespace std;

vector<int> data;
bool sort_comparator(const int &x, const int &y)
{
    return data[x] < data[y];
}

class Solution {
public:
    vector<int> twoSum(vector<int> &numbers, int target) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        result.clear();
        index.clear();
        data.clear();
        data = numbers;
        
        int i, j, n;
        
        n = numbers.size();
        if(n < 2){
            return result;
        }
        for(i = 0; i < n; ++i){
            index.push_back(i);
        }
        sort(index.begin(), index.end(), sort_comparator);
        
        int ll, rr, mm;
        for(i = 0; i < n; ++i){
            ll = i + 1;
            rr = n - 1;
            if(ll > rr || target - data[index[i]] < data[index[ll]] || target - data[index[i]] > data[index[rr]]){
                continue;
            }
            while(ll <= rr){
                mm = (ll + rr) / 2;
                if(target - data[index[i]] < data[index[mm]]){
                    rr = mm - 1;
                }else if(target - data[index[i]] > data[index[mm]]){
                    ll = mm + 1;
                }else{
                    if(index[i] < index[mm]){
                        result.push_back(index[i] + 1);
                        result.push_back(index[mm] + 1);
                    }else{
                        result.push_back(index[mm] + 1);
                        result.push_back(index[i] + 1);
                    }
                    break;
                }
            }
            if(ll <= rr){
                break;
            }
        }
        
        index.clear();
        data.clear();
        return result;
    }
private:
    vector<int> result;
    vector<int> index;
};