[LeetCode] 839. Similar String Groups
grandyang opened this issue · 0 comments
Two strings X
and Y
are similar if we can swap two letters (in different positions) of X
, so that it equals Y
.
For example, "tars"
and "rats"
are similar (swapping at positions 0
and 2
), and "rats"
and "arts"
are similar, but "star"
is not similar to "tars"
, "rats"
, or "arts"
.
Together, these form two connected groups by similarity: {"tars", "rats", "arts"}
and {"star"}
. Notice that "tars"
and "arts"
are in the same group even though they are not similar. Formally, each group is such that a word is in the group if and only if it is similar to at least one other word in the group.
We are given a list A
of strings. Every string in A
is an anagram of every other string in A
. How many groups are there?
Example 1:
Input: ["tars","rats","arts","star"]
Output: 2
Note:
A.length <= 2000
A[i].length <= 1000
A.length * A[i].length <= 20000
- All words in
A
consist of lowercase letters only. - All words in
A
have the same length and are anagrams of each other. - The judging time limit has been increased for this question.
这道题定义了字符串之间的一种相似关系,说是对于字符串X和Y,交换X中两个不同位置上的字符,若可以得到Y的话,就说明X和Y是相似的。现在给了我们一个字符串数组,要将相似的字符串放到一个群组里,这里同一个群组里的字符串不必任意两个都相似,而是只要能通过某些结点最终连着就行了,有点像连通图的感觉,将所有连通的结点算作一个群组,问整个数组可以分为多少个群组。由于这道题的本质就是求连通图求群组个数,既然是图,考察的就是遍历啦,就有 DFS 和 BFS 的解法。先来看 DFS 的解法,虽说本质是图的问题,但并不是真正的图,没有邻接链表啥的,这里判断两个结点是否相连其实就是判断是否相似。所以可以写一个判断是否相似的子函数,实现起来也非常的简单,只要按位置对比字符,若不相等则 diff 自增1,若 diff 大于2了直接返回 false,因为只有 diff 正好等于2或者0的时候才相似。题目中说了字符串之间都是异构词,说明字符的种类个数都一样,只是顺序不同,就不可能出现奇数的 diff,而两个字符串完全相等时也是满足要求的,是相似的。下面来进行 DFS 遍历,用一个 HashSet 来记录遍历过的字符串,对于遍历到的字符串,若已经在 HashSet 中存在了,直接跳过,否则结果 res 自增1,并调用递归函数。这里递归函数的作用是找出所有相似的字符串,首先还是判断当前字符串 str 是否访问过,是的话直接返回,否则加入 HashSet 中。然后再遍历一遍原字符串数组,每一个遍历到的字符串 word 都和 str 检测是否相似,相似的话就对这个 word 调用递归函数,这样就可以找出所有相似的字符串啦,参见代码如下:
解法一:
class Solution {
public:
int numSimilarGroups(vector<string>& A) {
int res = 0, n = A.size();
unordered_set<string> visited;
for (string str : A) {
if (visited.count(str)) continue;
++res;
helper(A, str, visited);
}
return res;
}
void helper(vector<string>& A, string& str, unordered_set<string>& visited) {
if (visited.count(str)) return;
visited.insert(str);
for (string word : A) {
if (isSimilar(word, str)) {
helper(A, word, visited);
}
}
}
bool isSimilar(string& str1, string& str2) {
for (int i = 0, cnt = 0; i < str1.size(); ++i) {
if (str1[i] == str2[i]) continue;
if (++cnt > 2) return false;
}
return true;
}
};
我们也可以使用 BFS 遍历来做,用一个 bool 型数组来标记访问过的单词,同时用队列 queue 来辅助计算。遍历所有的单词,假如已经访问过了,则直接跳过,否则就要标记为 true,然后结果 res 自增1,这里跟上面 DFS 的解法原理一样,要一次找完和当前结点相连的所有结点,只不过这里用了迭代的 BFS 的写法。先将当前字符串加入队列 queue 中,然后进行 while 循环,取出队首字符串,再遍历一遍所有字符串,遇到访问过的就跳过,然后统计每个字符串和队首字符串之间的不同字符个数,假如最终 diff 为0的话,说明是一样的,此时不加入队列,但是要标记这个字符串为 true;若最终 diff 为2,说明是相似的,除了要标记字符串为 true,还要将其加入队列进行下一轮查找,参见代码如下:
解法二:
class Solution {
public:
int numSimilarGroups(vector<string>& A) {
int res = 0, n = A.size();
vector<bool> visited(n);
queue<string> q;
for (int i = 0; i < n; ++i) {
if (visited[i]) continue;
visited[i] = true;
++res;
q.push(A[i]);
while (!q.empty()) {
string t = q.front(); q.pop();
for (int j = 0; j < n; ++j) {
if (visited[j]) continue;
int diff = 0;
for (int k = 0; k < A[j].size(); ++k) {
if (t[k] == A[j][k]) continue;
if (++diff > 2) break;
}
if (diff == 0) visited[j] = true;
if (diff == 2) {
visited[j] = true;
q.push(A[j]);
}
}
}
}
return res;
}
};
对于这种群组归类问题,很适合使用联合查找 Union Find 来做,LeetCode 中也有其他用到这个思路的题目,比如 Friend Circles,Accounts Merge,Redundant Connection II,Redundant Connection,Number of Islands II,Graph Valid Tree,和 Number of Connected Components in an Undirected Graph。都是要用一个 root 数组,每个点开始初始化为不同的值,如果两个点属于相同的组,就将其中一个点的 root 值赋值为另一个点的位置,这样只要是相同组里的两点,通过 getRoot 函数得到相同的值。所以这里对于每个结点 A[i],都遍历前面所有结点 A[j],假如二者不相似,直接跳过;否则将 A[j] 结点的 root 值更新为i,这样所有相连的结点的 root 值就相同了,一个群组中只有一个结点的 root 值会保留为其的初始值,所以最后只要统计到底还有多少个结点的 root 值还是初始值,就知道有多少个群组了,参见代码如下:
解法三:
class Solution {
public:
int numSimilarGroups(vector<string>& A) {
int res = 0, n = A.size();
vector<int> root(n);
for (int i = 0; i < n; ++i) root[i] = i;
for (int i = 1; i < n; ++i) {
for (int j = 0; j < i; ++j) {
if (!isSimilar(A[i], A[j])) continue;
root[getRoot(root, j)] = i;
}
}
for (int i = 0; i < n; ++i) {
if (root[i] == i) ++res;
}
return res;
}
int getRoot(vector<int>& root, int i) {
return (root[i] == i) ? i : getRoot(root, root[i]);
}
bool isSimilar(string& str1, string& str2) {
for (int i = 0, cnt = 0; i < str1.size(); ++i) {
if (str1[i] == str2[i]) continue;
if (++cnt > 2) return false;
}
return true;
}
};
Github 同步地址:
类似题目:
Number of Connected Components in an Undirected Graph
参考资料:
https://leetcode.com/problems/similar-string-groups/
https://leetcode.com/problems/similar-string-groups/discuss/200934/My-Union-Find-Java-Solution