专题九——哈希表

news2024/11/5 1:20:32

0简介

1两数之和

2判定是否互为字符重排

3存在重复元素

4存在重复元素 II

5字母异位词分组

0简介

1两数之和

oj链接：两数之和

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

//解法2
class Solution
{
public:
    vector<int> twoSum(vector<int> &nums, int target)
    {
        unordered_map<int, int> hash;
        for (int i = 0; i < nums.size(); i++)
        {
            if (hash.find(target - nums[i]) != hash.end())
            {
                return {hash[target - nums[i]], i};
            }
            hash[nums[i]] = i;
        }
        return {0, 0};
    }
};

2判定是否互为字符重排

oj链接：判定是否互为字符重排

//解法1
class Solution
{
public:
    bool CheckPermutation(string s1, string s2)
    {
        if (s1.size() != s2.size())
            return false;
        int hash1[26] = {0}, hash2[26] = {0};
        for (auto &s : s1)
            hash1[s - 'a']++;
        for (auto &s : s2)
            hash2[s - 'a']++;
        for (int i = 0; i < 26; i++)
        {
            if (hash1[i] != hash2[i])
                return false;
        }
        return true;
    }
};

//优化
class Solution
{
public:
    bool CheckPermutation(string s1, string s2)
    {
        if (s1.size() != s2.size())
            return false;
        int hash[26] = {0};
        for (auto &s : s1)
            hash[s - 'a']++;
        for (auto &s : s2)
        {
            if ((--hash[s - 'a']) < 0)
                return false;
        }
        return true;
    }
};

3存在重复元素

oj链接：存在重复元素

创建hash<数值，下标>：边存边判断即可

class Solution
{
public:
    bool containsDuplicate(vector<int> &nums)
    {
        unordered_map<int, int> hash;
        for (auto &num : nums)
        {
            if ((++hash[num]) == 2)
                return true;
        }
        return false;
    }
};

4存在重复元素 II

oj链接：存在重复元素 II

创建hash<数值下标> 进行存储时判断hash里有没有值；如果有将进行判断abs(i-j)<=k；没有进行存储(如果已经储存过了将其覆盖掉：因为判断为true时要两个值越靠近才有可能满足)

// 暴力：三层for循环
class Solution
{
public:
    bool containsNearbyDuplicate(vector<int> &nums, int k)
    {
        unordered_map<int, vector<int>> hash;
        for (int i = 0; i < nums.size(); i++)
            hash[nums[i]].push_back(i);
        for (int i = 0; i < nums.size(); i++)
        {
            vector<int> tmp = hash[nums[i]];
            if (tmp.size() >= 2)
            {
                for (int i = 0; i < tmp.size(); i++)
                {
                    for (int j = i + 1; j < tmp.size(); j++)
                    {
                        if (abs(tmp[i] - tmp[j]) <= k)
                            return true;
                    }
                }
            }
        }
        return false;
    }
};

// 优化
class Solution
{
public:
    bool containsNearbyDuplicate(vector<int> &nums, int k)
    {
        unordered_map<int, int> hash;
        for (int i = 0; i < nums.size(); i++)
        {
            if (hash.count(nums[i]) && abs(i - hash[nums[i]]) <= k)
                return true;
            hash[nums[i]] = i; // 覆盖不影响结果
        }
        return false;
    }
};

5字母异位词分组

oj链接：字母异位词分组

class Solution
{
public:
    vector<vector<string>> groupAnagrams(vector<string> &strs)
    {
        unordered_map<string, vector<string>> hash;
        for (auto &str : strs)
        {
            // 排序
            string tmp = str;
            sort(tmp.begin(), tmp.end());
            hash[tmp].push_back(str);
        }
        vector<vector<string>> ret;
        for (auto &[x, y] : hash)
        {
            ret.push_back(y);
        }
        return ret;
    }
};