LeetCode 1：两数之和（Two Sum）解法汇总

更多LeetCode题解

暴力解法/我的解法

这种办法很容易理解，就不赘述了，直接上代码

首先上java

public int[] twoSum(int[] nums, int target) {
    for (int i = 0; i < nums.length; i++) {
        for (int j = i + 1; j < nums.length; j++) {
            if (nums[j] == target - nums[i]) {
                return new int[] { i, j };
            }
        }
    }
    throw new IllegalArgumentException("No two sum solution");
}

然后是C++

class Solution {
public:
	vector<int> twoSum(vector<int>& nums, int target) {
		vector<int> result;
		vector<int>::iterator ib = nums.begin();
		vector<int>::iterator ie = nums.end();
		for (; ib != ie - 1; ib++)
		{
			vector<int>::iterator it = ib + 1;
			for (; it != ie; it++)
			{
				if (target == (*ib + *it))
				{
					result.push_back(ib - nums.begin());
					result.push_back(it - nums.begin());
					break;
				}
			}
		}
		return result;
	}
};

两遍哈希表

暴力法的时间复杂度很高，因此我们可以使用哈希表，两次循环。

首先还是java

public int[] twoSum(int[] nums, int target) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        map.put(nums[i], i);
    }
    for (int i = 0; i < nums.length; i++) {
        int complement = target - nums[i];
        if (map.containsKey(complement) && map.get(complement) != i) {
            return new int[] { i, map.get(complement) };
        }
    }
    throw new IllegalArgumentException("No two sum solution");
}

然后是C++

class Solution
{
public:
	vector<int> twoSum(vector<int> &nums, int target)
	{
		unordered_map<int, int> m;
		vector<int> result;
		for (size_t i = 0; i < nums.size(); ++i)
		{
			m[nums[i]] = i;
		}
		for (size_t i = 0; i < nums.size(); ++i)
		{
			int t = target - nums[i];
			if (m.count(t) && m[t] != i)
			{
				result.push_back(i);
				result.push_back(m[t]);
				break;
			}
		}
		return result;
	}
};

一遍哈希表

两遍哈希表可以简化为一遍

首先是java

public int[] twoSum(int[] nums, int target) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        int complement = target - nums[i];
        if (map.containsKey(complement)) {
            return new int[] { map.get(complement), i };
        }
        map.put(nums[i], i);
    }
    throw new IllegalArgumentException("No two sum solution");
}

然后是C++

class Solution
{
public:
	vector<int> twoSum(vector<int> &nums, int target)
	{
		unordered_map<int, int> m;
		vector<int> result;
		for (unsigned i = 0; i < nums.size(); i++)
		{
			if (m.count(target - nums[i]))
			{
				result.push_back(i);
				result.push_back(m[target - nums[i]]);
				break;
			}
			m[nums[i]] = i;
		}
		return result;
	}
};

拓展视野再来看看python

class Solution:
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        dict = {}
        for i in range(len(nums)):
            if target-nums[i] not in dict:
                dict[nums[i]]=i
            else:
                return [dict[target-nums[i]],i]