1. Two Sum【easy】

1. Two Sum【easy】

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].

解法一：

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

map搞一把

解法二：

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int, int> map;
        int n = (int)nums.size();
        for (int i = 0; i < n; i++) {
            auto p = map.find(target - nums[i]);
            if (p != map.end()) {
                return {p->second, i};
            }
            map[nums[i]] = i;
        }
    }
};

比较简洁的写法

扩展见：

167. Two Sum II - Input array is sorted【easy】

170. Two Sum III - Data structure design【easy】