Question

990. 等式方程的可满足性

English Version

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

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

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];
  }
}

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];
  }
};

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
}

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;
}