半路出家c++，急需提升c++熟练度，所以开始用c++重新刷leetcode。
 注重做题中c++基础的总结~

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].

my solution

刚开始没有注意到返回的是数组下标，直接排序了，后来在原基础上修改，直接复制了个原数组，遍历原数组，找结果在原来数组的位置。当然用hashmap代码更简洁。

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        int p1 = 0;
        int p2 = nums.size() - 1;
        vector<int> result;
        bool find_result = false;
        vector<int> old_nums(nums);
        sort(nums.begin(), nums.end());
        while (p1 < p2) {
            int sum = nums[p1] + nums[p2];
            if (sum > target) 
                p2--;
            else if (sum < target)
                p1++;
            else {        
                for (int i = 0; i < old_nums.size() ; i++) {
                    if (old_nums[i] == nums[p1]){
                        result.push_back(i);
                        continue;
                    }
                    if (old_nums[i] == nums[p2]){
                        result.push_back(i);
                        continue;
                    }
                }
                find_result = true;
                break;
            }
        }
    return result;
    }
};

best solution

#即hashmap
vector<int> twoSum(vector<int> &numbers, int target)
{
    //Key is the number and value is its index in the vector.
    unordered_map<int, int> hash;
    vector<int> result;
    for (int i = 0; i < numbers.size(); i++) {
        int numberToFind = target - numbers[i];

            //if numberToFind is found in map, return them
        if (hash.find(numberToFind) != hash.end()) {
                    //+1 because indices are NOT zero based
            result.push_back(hash[numberToFind] + 1);
            result.push_back(i + 1);            
            return result;
        }

            //number was not found. Put it in the map.
        //重载了[]可直接访问，插入。  operator[]在主键不存在时，自动创建
        hash[numbers[i]] = i;
        #hash.insert(pair<int, int>(numbers[i], i));
    }
    return result;
}

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int, int> lookup;
        for (int i = 0; i < nums.size(); ++i) {
            //使用count代替find
            if (lookup.count(target - nums[i])) {
                return {lookup[target - nums[i]], i};
            }
            lookup[nums[i]] = i;
        }
        return {};
    }
};

point

• vector排序：

#include<algorithm>
sort(v.begin(),v.end());

• vector 拷贝

vector<int> list;
#拷贝构造函数
vector<int> tem1(list);
#assign
vector<int> tem2;
tem2.assign(list.begin(), list.end());
#insert
vector<int> tem3;
tem3.insert(tem3.end(), list.begin(), list.end());

• unorderd_map

wiki:http://classfoo.com/ccby/arti...
与map区别在于 map会根据key进行排序：https://www.cnblogs.com/NeilZ...