6767
6868** 方法一:BFS**
6969
70+ 我们先根据二维数组 $edges$ 构建邻接表 $g,ドル其中 $g[ i] $ 表示节点 $i$ 的所有后继节点。
71+ 72+ 然后我们从小到大枚举节点 $i$ 作为祖先节点,使用 BFS 搜索节点 $i$ 的所有后继节点,把节点 $i$ 加入这些后继节点的祖先列表中。
73+ 74+ 时间复杂度 $O(n^2),ドル空间复杂度 $O(n^2)$。其中 $n$ 是节点数。
75+ 7076<!-- tabs:start -->
7177
7278### ** Python3**
7682``` python
7783class Solution :
7884 def getAncestors (self n : int , edges : List[List[int ]]) -> List[List[int ]]:
85+ def bfs (s : int ):
86+ q = deque([s])
87+ vis = {s}
88+ while q:
89+ i = q.popleft()
90+ for j in g[i]:
91+ if j not in vis:
92+ vis.add(j)
93+ q.append(j)
94+ ans[j].append(s)
95+ 7996 g = defaultdict(list )
8097 for u, v in edges:
81- g[v ].append(u )
82- ans = []
98+ g[u ].append(v )
99+ ans = [[] for _ in range (n) ]
83100 for i in range (n):
84- if not g[i]:
85- ans.append([])
86- continue
87- q = deque([i])
88- vis = [False ] * n
89- vis[i] = True
90- t = []
91- while q:
92- for _ in range (len (q)):
93- v = q.popleft()
94- for u in g[v]:
95- if not vis[u]:
96- vis[u] = True
97- q.append(u)
98- t.append(u)
99- ans.append(sorted (t))
101+ bfs(i)
100102 return ans
101103```
102104
@@ -106,39 +108,42 @@ class Solution:
106108
107109``` java
108110class Solution {
111+ private int n;
112+ private List<Integer > [] g;
113+ private List<List<Integer > > ans;
114+ 109115 public List<List<Integer > > getAncestors (int n , int [][] edges ) {
110- List<Integer > [] g = new List [n];
111- Arrays . setAll(g, k - > new ArrayList<> ());
112- for (int [] e : edges) {
113- g[e[1 ]]. add(e[0 ]);
116+ g = new List [n];
117+ this . n = n;
118+ Arrays . setAll(g, i - > new ArrayList<> ());
119+ for (var e : edges) {
120+ g[e[0 ]]. add(e[1 ]);
114121 }
115- List< List< Integer > > = new ArrayList<> ();
122+ ans = new ArrayList<> ();
116123 for (int i = 0 ; i < n; ++ i) {
117- List<Integer > t = new ArrayList<> ();
118- if (g[i]. isEmpty()) {
119- ans. add(t);
120- continue ;
121- }
122- Deque<Integer > q = new ArrayDeque<> ();
123- q. offer(i);
124- boolean [] vis = new boolean [n];
125- vis[i] = true ;
126- while (! q. isEmpty()) {
127- for (int j = q. size(); j > 0 ; -- j) {
128- int v = q. poll();
129- for (int u : g[v]) {
130- if (! vis[u]) {
131- vis[u] = true ;
132- q. offer(u);
133- t. add(u);
134- }
135- }
124+ ans. add(new ArrayList<> ());
125+ }
126+ for (int i = 0 ; i < n; ++ i) {
127+ bfs(i);
128+ }
129+ return ans;
130+ }
131+ 132+ private void bfs (int s ) {
133+ Deque<Integer > q = new ArrayDeque<> ();
134+ q. offer(s);
135+ boolean [] vis = new boolean [n];
136+ vis[s] = true ;
137+ while (! q. isEmpty()) {
138+ int i = q. poll();
139+ for (int j : g[i]) {
140+ if (! vis[j]) {
141+ vis[j] = true ;
142+ q. offer(j);
143+ ans. get(j). add(s);
136144 }
137145 }
138- Collections . sort(t);
139- ans. add(t);
140146 }
141- return ans;
142147 }
143148}
144149```
@@ -149,32 +154,31 @@ class Solution {
149154class Solution {
150155public:
151156 vector<vector<int >> getAncestors(int n, vector<vector<int >>& edges) {
152- vector<vector<int >> g(n);
153- for (auto& e : edges) g[ e[ 1]] .push_back(e[ 0] );
154- vector<vector<int >> ans;
155- for (int i = 0; i < n; ++i) {
156- vector<int > t;
157- if (g[ i] .empty()) {
158- ans.push_back(t);
159- continue;
160- }
161- queue<int > q{{i}};
162- vector<bool > vis(n);
163- vis[ i] = true;
164- while (!q.empty()) {
165- for (int j = q.size(); j > 0; --j) {
166- int v = q.front();
167- q.pop();
168- for (int u : g[ v] ) {
169- if (vis[ u] ) continue;
170- vis[ u] = true;
171- q.push(u);
172- t.push_back(u);
157+ vector<int > g[ n] ;
158+ for (auto& e : edges) {
159+ g[ e[ 0]] .push_back(e[ 1] );
160+ }
161+ vector<vector<int >> ans(n);
162+ auto bfs = [ &] (int s) {
163+ queue<int > q;
164+ q.push(s);
165+ bool vis[ n] ;
166+ memset(vis, 0, sizeof(vis));
167+ vis[ s] = true;
168+ while (q.size()) {
169+ int i = q.front();
170+ q.pop();
171+ for (int j : g[ i] ) {
172+ if (!vis[ j] ) {
173+ vis[ j] = true;
174+ ans[ j] .push_back(s);
175+ q.push(j);
173176 }
174177 }
175178 }
176- sort(t.begin(), t.end());
177- ans.push_back(t);
179+ };
180+ for (int i = 0; i < n; ++i) {
181+ bfs(i);
178182 }
179183 return ans;
180184 }
@@ -187,33 +191,27 @@ public:
187191func getAncestors(n int, edges [][]int) [][]int {
188192 g := make([][]int, n)
189193 for _, e := range edges {
190- g[e[1 ]] = append(g[e[1 ]], e[0 ])
194+ g[e[0 ]] = append(g[e[0 ]], e[1 ])
191195 }
192- var ans [][]int
193- for i := 0; i < n; i++ {
194- var t []int
195- if len(g[i]) == 0 {
196- ans = append(ans, t)
197- continue
198- }
199- q := []int{i}
196+ ans := make([][]int, n)
197+ bfs := func(s int) {
198+ q := []int{s}
200199 vis := make([]bool, n)
201- vis[i ] = true
200+ vis[s ] = true
202201 for len(q) > 0 {
203- for j := len(q); j > 0; j-- {
204- v := q[0]
205- q = q[1:]
206- for _, u := range g[v] {
207- if !vis[u] {
208- vis[u] = true
209- q = append(q, u)
210- t = append(t, u)
211- }
202+ i := q[0]
203+ q = q[1:]
204+ for _, j := range g[i] {
205+ if !vis[j] {
206+ vis[j] = true
207+ q = append(q, j)
208+ ans[j] = append(ans[j], s)
212209 }
213210 }
214211 }
215- sort.Ints(t)
216- ans = append(ans, t)
212+ }
213+ for i := 0; i < n; i++ {
214+ bfs(i)
217215 }
218216 return ans
219217}
@@ -222,7 +220,32 @@ func getAncestors(n int, edges [][]int) [][]int {
222220### ** TypeScript**
223221
224222``` ts
225- 223+ function getAncestors(n : number , edges : number [][]): number [][] {
224+ const : number [][] = Array .from ({ length: n }, () => []);
225+ for (const of edges ) {
226+ g [u ].push (v );
227+ }
228+ const : number [][] = Array .from ({ length: n }, () => []);
229+ const = (s : number ) => {
230+ const : number [] = [s ];
231+ const : boolean [] = Array .from ({ length: n }, () => false );
232+ vis [s ] = true ;
233+ while (q .length ) {
234+ const = q .shift ()! ;
235+ for (const of g [i ]) {
236+ if (! vis [j ]) {
237+ vis [j ] = true ;
238+ ans [j ].push (s );
239+ q .push (j );
240+ }
241+ }
242+ }
243+ };
244+ for (let i = 0 ; i < n ; ++ i ) {
245+ bfs (i );
246+ }
247+ return ans ;
248+ }
226249```
227250
228251### ** ...**
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