LeetCode题解 C++ Two Sum

最近要开始准备实习了,为了让自己显得不那么弱,所以拾起了以前的A题的感觉,开始在大名鼎鼎的LeetCode上开始刷题,废话不多说了,就从第一题开始吧。

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].

说实话重新开始做算法题已经手生了,刚开始做没有想很多,就开始暴力的用O(n^2)的方法,果然不出所料超时了.
原来思路是,对原来的数组做一个排序,然后两边的数往中间夹逼.代码如下

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;
    }
};

思考了一会才想到,原来的hash方法其实可以把时间缩短到O(n).
具体思路是我把原来的数组用map记录下来,每次记录时比较值和target的差值,如果在map中可以找到,就实现匹配,缩短时间.

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;
    }
};

加油继续A题……/(ㄒoㄒ)/~~