最近要開始準備實習了,為了讓自己顯得不那麼弱,所以拾起了以前的A題的感覺,開始在大名鼎鼎的LeetCode上開始刷題,廢話不多說了,就從第一題開始吧。

1. Two Sum

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].

說實話重新開始做演算法題已經手生了,剛開始做沒有想很多,就開始暴力的用O(n^2)的方法,果然不出所料超時了.
原來思路是,對原來的陣列做一個排序,然後兩邊的數往中間夾逼.程式碼如下

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector
 
<int> numbers = nums;
        std::sort(numbers.begin(), numbers.end());

        int length = numbers.size();
        int left = 0;
        int right = length - 1;
        int sum = 0;

        vector<int> index;

        while(left < right)
        {
            sum = numbers[left] + numbers[right];

            if
 
(sum == target)
            {
                for(int i = 0; i < length; ++ i)
                {
                    if(nums[i] == numbers[left])
                    {
                        index.push_back(i + 1);
                    }
                    else if(nums[i] == numbers[right])
                    {   
                        index.push_back(i + 1);
                    }
                    if(index.size() == 2)
                        break;
                }
            }
            else if(sum > target)
            {
                right --;
            }
            else
            {
                left ++;
            }
        }

        return index;
    }
};

思考了一會才想到,原來的hash方法其實可以把時間縮短到O(n).
具體思路是我把原來的陣列用map記錄下來,每次記錄時比較值和target的差值,如果在map中可以找到,就實現匹配,縮短時間.

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        map<int, int> m;
        vector<int> index;

        for(int i = 0; i < nums.size(); ++ i)
        {
            if(m.count(target - nums[i]) != 0)
            {
                index.push_back(m[target - nums[i]]);
                index.push_back(i);
                break;
            }
            m[nums[i]] = i;
        }

        return index;
    }
};

加油繼續A題……/(ㄒoㄒ)/~~