目錄

• 題目鏈接
• 註意點
• 解法
• 小結

題目鏈接

Satisfiability of Equality Equations - LeetCode

註意點

• 必須要初始化pre

解法

解法一：典型的並查集算法應用。先遍歷所有等式，將等號兩邊的字母加入同一分類，每類中的字母都是相等的。然後遍歷不等式，如果不等號兩邊的字母屬於同一類則返回false。時間復雜度O(nm)

class Solution {
public:
    map<char,char> pre;
    char Find(char x)
    {
        char r = x;
        while(pre[r] != r)
        {
            r = pre[r];
        }
        return r;
    }
    void Join(char x,char y)
    {
        char fx = Find(x);
        char fy = Find(y);
        if(fx != fy) pre[fx] = fy;
    }
    void Init()
    {
        char letter[26] ={'a','b','c','d','e','f','g','h','i',
                          'j','k','l','m','n','o','p','q','r',
                          's','t','u','v','w','x','y','z'};
        for (auto l:letter)
        {
            pre[l] = l;
        }
        
    }
    bool equationsPossible(vector<string>& equations) {
        Init();
        for(auto e:equations)
        {
            if(e[1] == '=') Join(e[0],e[3]);
        }
        for(auto e:equations)
        {
            if(e[1] == '!' && Find(e[0]) == Find(e[3]))
            {
                return false;
            }
        }
        return true;
    }
};
 

小結

• 快速搞定並查集算法

Satisfiability of Equality Equations - LeetCode