Two Sum-----LeetCode

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

 

解題思路：

（1）O(nlogn)。排序，然后兩個指針一前一后。因為題中說明了只有一對答案，因此不需要考慮重復的情況。

（2）O(n)。哈希表。將每個數字放在map中，歷遍數組，如果出現和數組中的某一個值相加為target的時候，break。這個方法同樣適用於多組解的情況。

 

代碼：

struct Node
{
    int num, pos;
};
bool cmp(Node a, Node b)
{
    return a.num < b.num;
}
class Solution {
public:
    vector<int> twoSum(vector<int> &numbers, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<int> result;
        vector<Node> array;
        for (int i = 0; i < numbers.size(); i++)
        {
            Node temp;
            temp.num = numbers[i];
            temp.pos = i;
            array.push_back(temp);
        }

        sort(array.begin(), array.end(), cmp);
        for (int i = 0, j = array.size() - 1; i != j;)
        {
            int sum = array[i].num + array[j].num;
            if (sum == target)
            {
                if (array[i].pos < array[j].pos)
                {
                    result.push_back(array[i].pos + 1);
                    result.push_back(array[j].pos + 1);
                } else
                {
                    result.push_back(array[j].pos + 1);
                    result.push_back(array[i].pos + 1);
                }
                break;
            } else if (sum < target)
            {
                i++;
            } else if (sum > target)
            {
                j--;
            }
        }
        return result;
    }
};