LeetCode TwoSum

class Solution {

public:

    vector<int> twoSum(vector<int> &numbers, int target) {

        vector<int> ret;

        vector<pair<int, int> > nums;

        for (int i=0; i<numbers.size(); i++) {

            nums.push_back(make_pair(numbers[i], i+1));

        }

        sort(nums.begin(), nums.end());



        int i = 0, j = nums.size() - 1;

        int sum = 0;

        while(i<j) {

            sum = nums[i].first + nums[j].first;

            if (sum > target) {

                j--;

            } else if (sum < target) {

                i++;

            } else {

                break;

            }

        }

        ret.push_back(nums[i].second);

        ret.push_back(nums[j].second);

        sort(ret.begin(), ret.end());

        return ret;

    }

};

怒りが存在感を磨く
第2ラウンド:
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=9Output: index1=1, index2=2
初めて作ったのは少しざらざらしているような気がして、答えを見て簡潔になりました.

class Solution {

public:

    vector<int> twoSum(vector<int> &numbers, int target) {

        unordered_map<int, int> n2i;

        int len = numbers.size();

        vector<int> res;

        for (int i=0; i<len; i++) {

            int cur = numbers[i];

            if (n2i.count(target - cur) > 0) {

                res = {n2i[target-cur] + 1, i + 1};

                return res;

            }

            n2i.insert({cur, i});

        }

        return res;

    }

};