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;
    }
};

怒刷存在感

第二轮：

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

感觉第一次做的有点粗糙，看了一发答案简洁许多：

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;
    }
};