[leetcode] Two Sum

From: https://leetcode.com/problems/two-sum/

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

class Solution {public:    vector<int> twoSum(vector<int>& nums, int target) {vector<int> copy(nums.begin(), nums.end());sort(copy.begin(), copy.end());int st=0, ed=copy.size()-1;while(st < ed) {if(copy[st]+copy[ed] == target) break;if(copy[st]+copy[ed] > target) ed--;else st++;}vector<int> ans;if(copy[st]+copy[ed] != target) return ans;int t1=copy[st], t2=copy[ed];st=-1; ed=-1;for(int i=0, len=nums.size(); i<len; i++) {if(nums[i]==t1 || nums[i]==t2) {if(st == -1) st=i;else if(nums[st]+nums[i] == target){ ed=i; break;}}}ans.push_back(st+1);ans.push_back(ed+1);return ans;    }};

class Solution {public:    vector<int> twoSum(vector<int> &nums, int target) {        unordered_map<int,int> occured;        vector<int> ans;                for(int i=0; i<nums.size(); ++i) {            if(occured.find(target-nums[i])!=occured.end()) {                ans.push_back(occured[target-nums[i]]+1);                ans.push_back(i+1);                break;            }            if(occured.find(nums[i]) == occured.end()) {                occured[nums[i]] = i;            }        }                return ans;    }};

前者使用vector，后者使用了map，后者更占用空间；前者排先排序，用时O(nlgn)，后者直接找，O(n)。

貌似后者的速度应该更快，但从leetcode执行结果看，前者更快。