1 Two Sum 兩數(shù)之和

Description:
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].

題目描述:
給定一個整數(shù)數(shù)組 nums 和一個目標(biāo)值 target，請你在該數(shù)組中找出和為目標(biāo)值的那兩個整數(shù)，并返回他們的數(shù)組下標(biāo)。

你可以假設(shè)每種輸入只會對應(yīng)一個答案。但是，你不能重復(fù)利用這個數(shù)組中同樣的元素。

示例:

給定 nums = [2, 7, 11, 15], target = 9

因為 nums[0] + nums[1] = 2 + 7 = 9
所以返回 [0, 1]

思路:

1. 暴力法: 對每個數(shù), 遍歷剩下的數(shù)求是否存在解. 時間復(fù)雜度O(n^2), 空間復(fù)雜度O(1)
2. 一次遍歷哈希法: 建立一個map存放相應(yīng)元素對應(yīng)的目標(biāo)元素, 并在插入的時候檢查, 時間復(fù)雜度O(n), 空間復(fù)雜度O(n)

代碼:
C++:

class Solution 
{
public:
    vector<int> twoSum(vector<int>& nums, int target) 
    {
        vector<int> result;
        // key為nums的值, value為nums下標(biāo)
        map<int, int> int_map;
        for (int i = 0; i < nums.size(); i++) 
        {
            // map.count()測試主鍵是否存在, 若存在返回1
            if (int_map.count(nums[i]) != 0) 
            {
                // push_back(elem)在容器最后位置添加一個元素elem
                result.push_back(int_map[nums[i]]);
                result.push_back(i);
                break;
            }
            int_map[target- nums[i]] = i;
        }
        return result;     
    }
};

Java:

class Solution {
    public int[] twoSum(int[] nums, int target) {
        Map<Integer, Integer> map = new HashMap<>();
        int[] result = new int[2];
        for (int i = 0; i < nums.length; i++) {
            int complement = target - nums[i];
            if (map.containsKey(complement)) {
                result[0] = map.get(complement);
                result[1] = i;
                return result;
            }
            map.put(nums[i], i);
        }
        return result;
    }
}

Python:

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        dic = {}
        for index, num in enumerate(nums):
            complement = target - num
            if complement in dic:
                return [dic[complement], index]
            dic[num] = index