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feat: update lc problems (doocs#3679)
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20 files changed

+1813
-399
lines changed

20 files changed

+1813
-399
lines changed

‎solution/0600-0699/0685.Redundant Connection II/README.md

Lines changed: 610 additions & 111 deletions
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‎solution/0600-0699/0685.Redundant Connection II/README_EN.md

Lines changed: 613 additions & 110 deletions
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‎solution/0600-0699/0685.Redundant Connection II/Solution.cpp

Lines changed: 35 additions & 40 deletions
Original file line numberDiff line numberDiff line change
@@ -1,48 +1,43 @@
1-
class UnionFind {
2-
public:
3-
vector<int> p;
4-
int n;
5-
6-
UnionFind(int _n)
7-
: n(_n)
8-
, p(_n) {
9-
iota(p.begin(), p.end(), 0);
10-
}
11-
12-
bool unite(int a, int b) {
13-
int pa = find(a), pb = find(b);
14-
if (pa == pb) return false;
15-
p[pa] = pb;
16-
--n;
17-
return true;
18-
}
19-
20-
int find(int x) {
21-
if (p[x] != x) p[x] = find(p[x]);
22-
return p[x];
23-
}
24-
};
25-
261
class Solution {
272
public:
283
vector<int> findRedundantDirectedConnection(vector<vector<int>>& edges) {
294
int n = edges.size();
30-
vector<int> p(n + 1);
31-
for (int i = 0; i <= n; ++i) p[i] = i;
32-
UnionFind uf(n + 1);
33-
int conflict = -1, cycle = -1;
5+
vector<int> ind(n);
6+
for (const auto& e : edges) {
7+
++ind[e[1] - 1];
8+
}
9+
vector<int> dup;
3410
for (int i = 0; i < n; ++i) {
35-
int u = edges[i][0], v = edges[i][1];
36-
if (p[v] != v)
37-
conflict = i;
38-
else {
39-
p[v] = u;
40-
if (!uf.unite(u, v)) cycle = i;
11+
if (ind[edges[i][1] - 1] == 2) {
12+
dup.push_back(i);
4113
}
4214
}
43-
if (conflict == -1) return edges[cycle];
44-
int v = edges[conflict][1];
45-
if (cycle != -1) return {p[v], v};
46-
return edges[conflict];
15+
vector<int> p(n);
16+
iota(p.begin(), p.end(), 0);
17+
function<int(int)> find = [&](int x) {
18+
return x == p[x] ? x : p[x] = find(p[x]);
19+
};
20+
if (!dup.empty()) {
21+
for (int i = 0; i < n; ++i) {
22+
if (i == dup[1]) {
23+
continue;
24+
}
25+
int pu = find(edges[i][0] - 1);
26+
int pv = find(edges[i][1] - 1);
27+
if (pu == pv) {
28+
return edges[dup[0]];
29+
}
30+
p[pu] = pv;
31+
}
32+
return edges[dup[1]];
33+
}
34+
for (int i = 0;; ++i) {
35+
int pu = find(edges[i][0] - 1);
36+
int pv = find(edges[i][1] - 1);
37+
if (pu == pv) {
38+
return edges[i];
39+
}
40+
p[pu] = pv;
41+
}
4742
}
48-
};
43+
};

‎solution/0600-0699/0685.Redundant Connection II/Solution.go

Lines changed: 36 additions & 48 deletions
Original file line numberDiff line numberDiff line change
@@ -1,57 +1,45 @@
1-
type unionFind struct {
2-
p []int
3-
n int
4-
}
5-
6-
func newUnionFind(n int) *unionFind {
7-
p := make([]int, n)
8-
for i := range p {
9-
p[i] = i
10-
}
11-
return &unionFind{p, n}
12-
}
13-
14-
func (uf *unionFind) find(x int) int {
15-
if uf.p[x] != x {
16-
uf.p[x] = uf.find(uf.p[x])
17-
}
18-
return uf.p[x]
19-
}
20-
21-
func (uf *unionFind) union(a, b int) bool {
22-
if uf.find(a) == uf.find(b) {
23-
return false
24-
}
25-
uf.p[uf.find(a)] = uf.find(b)
26-
uf.n--
27-
return true
28-
}
29-
301
func findRedundantDirectedConnection(edges [][]int) []int {
312
n := len(edges)
32-
p := make([]int, n+1)
3+
ind := make([]int, n)
4+
for _, e := range edges {
5+
ind[e[1]-1]++
6+
}
7+
dup := []int{}
8+
for i, e := range edges {
9+
if ind[e[1]-1] == 2 {
10+
dup = append(dup, i)
11+
}
12+
}
13+
p := make([]int, n)
3314
for i := range p {
3415
p[i] = i
3516
}
36-
uf := newUnionFind(n + 1)
37-
conflict, cycle := -1, -1
38-
for i, e := range edges {
39-
u, v := e[0], e[1]
40-
if p[v] != v {
41-
conflict = i
42-
} else {
43-
p[v] = u
44-
if !uf.union(u, v) {
45-
cycle = i
46-
}
17+
var find func(int) int
18+
find = func(x int) int {
19+
if p[x] != x {
20+
p[x] = find(p[x])
4721
}
22+
return p[x]
4823
}
49-
if conflict == -1 {
50-
return edges[cycle]
24+
if len(dup) > 0 {
25+
for i, e := range edges {
26+
if i == dup[1] {
27+
continue
28+
}
29+
pu, pv := find(e[0]-1), find(e[1]-1)
30+
if pu == pv {
31+
return edges[dup[0]]
32+
}
33+
p[pu] = pv
34+
}
35+
return edges[dup[1]]
5136
}
52-
v := edges[conflict][1]
53-
if cycle != -1 {
54-
return []int{p[v], v}
37+
for _, e := range edges {
38+
pu, pv := find(e[0]-1), find(e[1]-1)
39+
if pu == pv {
40+
return e
41+
}
42+
p[pu] = pv
5543
}
56-
return edges[conflict]
57-
}
44+
return nil
45+
}

‎solution/0600-0699/0685.Redundant Connection II/Solution.java

Lines changed: 32 additions & 45 deletions
Original file line numberDiff line numberDiff line change
@@ -1,61 +1,48 @@
11
class Solution {
2+
private int[] p;
3+
24
public int[] findRedundantDirectedConnection(int[][] edges) {
35
int n = edges.length;
4-
int[] p = new int[n + 1];
5-
for (int i = 0; i <= n; ++i) {
6-
p[i] = i;
7-
}
8-
UnionFind uf = new UnionFind(n + 1);
9-
int conflict = -1, cycle = -1;
10-
for (int i = 0; i < n; ++i) {
11-
int u = edges[i][0], v = edges[i][1];
12-
if (p[v] != v) {
13-
conflict = i;
14-
} else {
15-
p[v] = u;
16-
if (!uf.union(u, v)) {
17-
cycle = i;
18-
}
19-
}
6+
int[] ind = new int[n];
7+
for (var e : edges) {
8+
++ind[e[1] - 1];
209
}
21-
if (conflict == -1) {
22-
return edges[cycle];
23-
}
24-
int v = edges[conflict][1];
25-
if (cycle != -1) {
26-
return new int[] {p[v], v};
27-
}
28-
return edges[conflict];
29-
}
30-
}
31-
32-
class UnionFind {
33-
public int[] p;
34-
public int n;
35-
36-
public UnionFind(int n) {
10+
List<Integer> dup = new ArrayList<>();
3711
p = new int[n];
3812
for (int i = 0; i < n; ++i) {
13+
if (ind[edges[i][1] - 1] == 2) {
14+
dup.add(i);
15+
}
3916
p[i] = i;
4017
}
41-
this.n = n;
42-
}
43-
44-
public boolean union(int a, int b) {
45-
int pa = find(a);
46-
int pb = find(b);
47-
if (pa == pb) {
48-
return false;
18+
if (!dup.isEmpty()) {
19+
for (int i = 0; i < n; ++i) {
20+
if (i == dup.get(1)) {
21+
continue;
22+
}
23+
int pu = find(edges[i][0] - 1);
24+
int pv = find(edges[i][1] - 1);
25+
if (pu == pv) {
26+
return edges[dup.get(0)];
27+
}
28+
p[pu] = pv;
29+
}
30+
return edges[dup.get(1)];
31+
}
32+
for (int i = 0;; ++i) {
33+
int pu = find(edges[i][0] - 1);
34+
int pv = find(edges[i][1] - 1);
35+
if (pu == pv) {
36+
return edges[i];
37+
}
38+
p[pu] = pv;
4939
}
50-
p[pa] = pb;
51-
--n;
52-
return true;
5340
}
5441

55-
public int find(int x) {
42+
private int find(int x) {
5643
if (p[x] != x) {
5744
p[x] = find(p[x]);
5845
}
5946
return p[x];
6047
}
61-
}
48+
}

‎solution/0600-0699/0685.Redundant Connection II/Solution.js

Lines changed: 44 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,44 @@
1+
/**
2+
* @param {number[][]} edges
3+
* @return {number[]}
4+
*/
5+
var findRedundantDirectedConnection = function (edges) {
6+
const n = edges.length;
7+
const ind = Array(n).fill(0);
8+
for (const [_, v] of edges) {
9+
++ind[v - 1];
10+
}
11+
const dup = [];
12+
for (let i = 0; i < n; ++i) {
13+
if (ind[edges[i][1] - 1] === 2) {
14+
dup.push(i);
15+
}
16+
}
17+
const p = Array.from({ length: n }, (_, i) => i);
18+
const find = x => {
19+
if (p[x] !== x) {
20+
p[x] = find(p[x]);
21+
}
22+
return p[x];
23+
};
24+
if (dup.length) {
25+
for (let i = 0; i < n; ++i) {
26+
if (i === dup[1]) {
27+
continue;
28+
}
29+
const [pu, pv] = [find(edges[i][0] - 1), find(edges[i][1] - 1)];
30+
if (pu === pv) {
31+
return edges[dup[0]];
32+
}
33+
p[pu] = pv;
34+
}
35+
return edges[dup[1]];
36+
}
37+
for (let i = 0; ; ++i) {
38+
const [pu, pv] = [find(edges[i][0] - 1), find(edges[i][1] - 1)];
39+
if (pu === pv) {
40+
return edges[i];
41+
}
42+
p[pu] = pv;
43+
}
44+
};

‎solution/0600-0699/0685.Redundant Connection II/Solution.py

Lines changed: 23 additions & 33 deletions
Original file line numberDiff line numberDiff line change
@@ -1,37 +1,27 @@
1-
class UnionFind:
2-
def __init__(self, n):
3-
self.p = list(range(n))
4-
self.n = n
5-
6-
def union(self, a, b):
7-
if self.find(a) == self.find(b):
8-
return False
9-
self.p[self.find(a)] = self.find(b)
10-
self.n -= 1
11-
return True
12-
13-
def find(self, x):
14-
if self.p[x] != x:
15-
self.p[x] = self.find(self.p[x])
16-
return self.p[x]
17-
18-
191
class Solution:
202
def findRedundantDirectedConnection(self, edges: List[List[int]]) -> List[int]:
3+
def find(x: int) -> int:
4+
if p[x] != x:
5+
p[x] = find(p[x])
6+
return p[x]
7+
218
n = len(edges)
22-
p = list(range(n + 1))
23-
uf = UnionFind(n + 1)
24-
conflict = cycle = None
9+
ind = [0] * n
10+
for _, v in edges:
11+
ind[v - 1] += 1
12+
dup = [i for i, (_, v) in enumerate(edges) if ind[v - 1] == 2]
13+
p = list(range(n))
14+
if dup:
15+
for i, (u, v) in enumerate(edges):
16+
if i == dup[1]:
17+
continue
18+
pu, pv = find(u - 1), find(v - 1)
19+
if pu == pv:
20+
return edges[dup[0]]
21+
p[pu] = pv
22+
return edges[dup[1]]
2523
for i, (u, v) in enumerate(edges):
26-
if p[v] != v:
27-
conflict = i
28-
else:
29-
p[v] = u
30-
if not uf.union(u, v):
31-
cycle = i
32-
if conflict is None:
33-
return edges[cycle]
34-
v = edges[conflict][1]
35-
if cycle is not None:
36-
return [p[v], v]
37-
return edges[conflict]
24+
pu, pv = find(u - 1), find(v - 1)
25+
if pu == pv:
26+
return edges[i]
27+
p[pu] = pv

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