990. Satisfiability of Equality Equations. 并查集
Given an array equations of strings that represent relationships between variables, each string equations[i] has length 4 and takes one of two different forms: "a==b" or "a!=b". Here, a and b are lowercase letters (not necessarily different) that represent one-letter variable names.
Return true if and only if it is possible to assign integers to variable names so as to satisfy all the given equations.
Example 1:
Input: ["a==b","b!=a"]
Output: false
Explanation: If we assign say, a = 1 and b = 1, then the first equation is satisfied, but not the second. There is no way to assign the variables to satisfy both equations.
Example 2:
Input: ["ba","ab"]
Output: true
Explanation: We could assign a = 1 and b = 1 to satisfy both equations.
Example 3:
Input: ["ab","bc","a==c"]
Output: true
Example 4:
Input: ["ab","b!=c","ca"]
Output: false
Example 5:
Input: ["cc","bd","x!=z"]
Output: true
Note:
1 <= equations.length <= 500
equations[i].length == 4
equations[i][0] and equations[i][3] are lowercase letters
equations[i][1] is either '=' or '!'
equations[i][2] is '='
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/satisfiability-of-equality-equations
直接并查集
class Solution {
public:
vector <int> fa;
int find(int x){
return x == fa[x] ? x : fa[x] = find(fa[x]);
}
void unite(int x,int y){
int xx = find(x);
int yy = find(y);
fa[xx] = yy;
}
bool same(int x,int y){
return find(x) == find(y);
}
bool equationsPossible(vector<string>& equations) {
fa.resize(26);
iota(fa.begin(), fa.end(), 0);
for (auto str : equations){
if (str[1] == '='){
int x = str[0] - 'a';
int y = str[3] - 'a';
unite(x,y);
}
}
for (auto str : equations){
if (str[1] == '!'){
int x = str[0] - 'a';
int y = str[3] - 'a';
if (same(x,y)){
return false;
}
}
}
return true;
}
};