LeetCode Solution: 1. Two Sum

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].

解析1：

class Solution {
public:
    std::vector<int> twoSum(std::vector<int>& nums, int target) 
    {
         for (size_t i = 0; i < nums.size(); ++i) 
            for(size_t j = 0; j < nums.size(); ++j)
                if (i != j && nums[i] + nums[j] == target)
                    return {(int)i, (int)j};
        return {-1, -1};
    }
};

十分暴力的写法，时间复杂度为O(n^2)，直接遍历容器，寻找到下标不同，并且相加等于target的下标，然后返回

解析2：

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

解析2的算法时间复杂度为O(n）理解解析2需要理解 map容器
map容器是一个存放关键字和值对的容器，关键字可以是任意类型，直接访问关键字下标可以得到对应的值，用到了C++STL里面的unordered_map
首先定义一个std::unordered_map<int, int> hash和 std::vector<int> result还有我们需要的数wanted_num
进入循环，

for (int i = 0; i < nums.size(); i++) {
            if (hash.find(wanted_num) != hash.end()) {
                result.push_back(hash[wanted_num]);
                result.push_back(i);
                return result;
            }
            hash[nums[i]] = i;
        }

访问nums的每一个元素，进入判定

1. 如果wanted_num在hash里的话，就把wanted_num的索引和当前的i放入result中，作为最后结果
2. 如果没有在hash表里面，当前的元素放入hash表里，并设置下标为i

最后一定会把wanted_num放进hash表中，即解

时间复杂度O(n)

---