839. Similar String Groups
If the exchange string Xof letters in two different positions, such as the string Yare equal, so called X, and Ytwo similar strings. If the two strings themselves are equal, then they are also similar.
For example, "tars"and "rats"similar (exchange 0and 2position); "rats"and "arts"also similar, but "star"does not "tars", "rats"or "arts"similar.
In short, they form two association groups through similarity: {"tars", "rats", "arts"}and {"star"}. Note that "tars"and "arts"are in the same group, even if they are not similar. Formally, for each group, to determine a word in the group, it only needs to be similar to at least one word in the group.
Give you a list of strings strs. Each string is a list of strsother letter of the word ectopic all strings. Ask strshow much of a string of similar groups?
Example 1:
输入：strs = ["tars","rats","arts","star"]
输出：2
Example 2:
输入：strs = ["omv","ovm"]
输出：1
prompt:
- 1 <= strs.length <= 100
- 1 <= strs[i].length <= 1000
- sum(strs[i].length) <= 2 * 104
- strs[i] contains only lowercase letters.
- All the words in strs have the same length and are diaphanous words of each other.
Remarks:
- Anagram is a new word formed by changing the position (order) of the letters of a string.
Method 1: Collect and check
Problem-solving ideas
The last day of January, of course, it must end with "Combined Check Collection"~~
There is no special solution, the two-layer loop honestly compares whether two strings are "similar strings". If the string is similar unionto the can together.
Reference Code
public int numSimilarGroups(String[] strs) {
int n = strs.length;
UnionFind unionFind = new UnionFind(n);
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
if (unionFind.connected(i, j)) {
continue;
}
if (isSimilar(strs[i], strs[j])) {
unionFind.union(i, j);
}
}
}
return unionFind.getCount();
}
// 判断两个字符串是否为“相似字符串”
public boolean isSimilar(String s1, String s2) {
char[] cs1 = s1.toCharArray(), cs2 = s2.toCharArray();
int num = 0;
for (int i = 0; i < cs1.length; i++) {
if (cs1[i] != cs2[i]) {
num++;
if (num > 2) {
return false;
}
}
}
return true;
}
// 并查集模板
class UnionFind {
private int[] parent;
private int count;
public UnionFind(int n) {
count = n;
parent = new int[n];
for (int i = 0; i < n; i++) {
parent[i] = i;
}
}
public void union(int x, int y) {
int rootX = find(x), rootY = find(y);
if (rootX == rootY) {
return;
}
parent[rootX] = rootY;
count--;
}
public int find(int x) {
return parent[x] == x ? parent[x] : (parent[x] = find(parent[x]));
}
public boolean connected(int x, int y) {
return find(x) == find(y);
}
public int getCount() {
return count;
}
}
Results of the
