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990. 等式方程的可满足性

题目描述

给定一个由表示变量之间关系的字符串方程组成的数组，每个字符串方程 equations[i] 的长度为 4，并采用两种不同的形式之一："a==b" 或 "a!=b"。在这里，a 和 b 是小写字母（不一定不同），表示单字母变量名。

只有当可以将整数分配给变量名，以便满足所有给定的方程时才返回 true，否则返回 false。 

 

示例 1：

输入：["a==b","b!=a"]
输出：false
解释：如果我们指定，a = 1 且 b = 1，那么可以满足第一个方程，但无法满足第二个方程。没有办法分配变量同时满足这两个方程。

示例 2：

输入：["b==a","a==b"]
输出：true
解释：我们可以指定 a = 1 且 b = 1 以满足满足这两个方程。

示例 3：

输入：["a==b","b==c","a==c"]
输出：true

示例 4：

输入：["a==b","b!=c","c==a"]
输出：false

示例 5：

输入：["c==c","b==d","x!=z"]
输出：true

 

提示：

  1. 1 <= equations.length <= 500
  2. equations[i].length == 4
  3. equations[i][0] 和 equations[i][3] 是小写字母
  4. equations[i][1] 要么是 '='，要么是 '!'
  5. equations[i][2] 是 '='

解法

方法一

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class Solution:
    def equationsPossible(self, equations: List[str]) -> bool:
        def find(x):
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        p = list(range(26))
        for e in equations:
            a, b = ord(e[0]) - ord('a'), ord(e[-1]) - ord('a')
            if e[1] == '=':
                p[find(a)] = find(b)
        for e in equations:
            a, b = ord(e[0]) - ord('a'), ord(e[-1]) - ord('a')
            if e[1] == '!' and find(a) == find(b):
                return False
        return True

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class Solution {
    private int[] p;

    public boolean equationsPossible(String[] equations) {
        p = new int[26];
        for (int i = 0; i < 26; ++i) {
            p[i] = i;
        }
        for (String e : equations) {
            int a = e.charAt(0) - 'a', b = e.charAt(3) - 'a';
            if (e.charAt(1) == '=') {
                p[find(a)] = find(b);
            }
        }
        for (String e : equations) {
            int a = e.charAt(0) - 'a', b = e.charAt(3) - 'a';
            if (e.charAt(1) == '!' && find(a) == find(b)) {
                return false;
            }
        }
        return true;
    }

    private int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
}

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class Solution {
public:
    vector<int> p;

    bool equationsPossible(vector<string>& equations) {
        p.resize(26);
        for (int i = 0; i < 26; ++i) p[i] = i;
        for (auto& e : equations) {
            int a = e[0] - 'a', b = e[3] - 'a';
            if (e[1] == '=') p[find(a)] = find(b);
        }
        for (auto& e : equations) {
            int a = e[0] - 'a', b = e[3] - 'a';
            if (e[1] == '!' && find(a) == find(b)) return false;
        }
        return true;
    }

    int find(int x) {
        if (p[x] != x) p[x] = find(p[x]);
        return p[x];
    }
};

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func equationsPossible(equations []string) bool {
    p := make([]int, 26)
    for i := 1; i < 26; i++ {
        p[i] = i
    }
    var find func(x int) int
    find = func(x int) int {
        if p[x] != x {
            p[x] = find(p[x])
        }
        return p[x]
    }
    for _, e := range equations {
        a, b := int(e[0]-'a'), int(e[3]-'a')
        if e[1] == '=' {
            p[find(a)] = find(b)
        }
    }
    for _, e := range equations {
        a, b := int(e[0]-'a'), int(e[3]-'a')
        if e[1] == '!' && find(a) == find(b) {
            return false
        }
    }
    return true
}

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class UnionFind {
    private parent: number[];

    constructor() {
        this.parent = Array.from({ length: 26 }).map((_, i) => i);
    }

    find(index: number) {
        if (this.parent[index] === index) {
            return index;
        }
        this.parent[index] = this.find(this.parent[index]);
        return this.parent[index];
    }

    union(index1: number, index2: number) {
        this.parent[this.find(index1)] = this.find(index2);
    }
}

function equationsPossible(equations: string[]): boolean {
    const uf = new UnionFind();
    for (const [a, s, _, b] of equations) {
        if (s === '=') {
            const index1 = a.charCodeAt(0) - 'a'.charCodeAt(0);
            const index2 = b.charCodeAt(0) - 'a'.charCodeAt(0);
            uf.union(index1, index2);
        }
    }
    for (const [a, s, _, b] of equations) {
        if (s === '!') {
            const index1 = a.charCodeAt(0) - 'a'.charCodeAt(0);
            const index2 = b.charCodeAt(0) - 'a'.charCodeAt(0);
            if (uf.find(index1) === uf.find(index2)) {
                return false;
            }
        }
    }
    return true;
}

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