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  • BFS vs DFS

    1 Clone Graph   1  copy ervery nodes by bfs  2  add neighbors

        public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) 
        {
            if (node == null) {
                return node;
            }
            
            List<UndirectedGraphNode> nodes = new ArrayList<>();
            Map<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<>();
            nodes.add(node);
            map.put(node, new UndirectedGraphNode(node.label));
            
            int start = 0;
            while (start < nodes.size()) {
                UndirectedGraphNode head = nodes.get(start++);
                for (int i = 0; i < head.neighbors.size(); i++) {
                    UndirectedGraphNode neighbor = head.neighbors.get(i);
                    if (!map.containsKey(neighbor)) {
                        nodes.add(neighbor);
                        map.put(neighbor, new UndirectedGraphNode(neighbor.label));
                    }
                }
            }
            
            for (int i = 0; i < nodes.size(); i++) {
                UndirectedGraphNode newNode = map.get(nodes.get(i));
                for (int j = 0; j < nodes.get(i).neighbors.size(); j++) {
                    newNode.neighbors.add(map.get(nodes.get(i).neighbors.get(j)));
                }
            }
            
            return map.get(node);
        }
    View Code

    2 Topological Sorting   1 store nodes and rudu  2 find nodes has 0 rudu  3 bfs

        public ArrayList<DirectedGraphNode> topSort(ArrayList<DirectedGraphNode> graph) {
            // write your code here
            ArrayList<DirectedGraphNode> result = new ArrayList<>();
            Map<DirectedGraphNode, Integer> map = new HashMap<>();
            
            for (DirectedGraphNode node : graph) {
                for (DirectedGraphNode neighbor : node.neighbors) {
                    if (map.containsKey(neighbor)) {
                        map.put(neighbor, map.get(neighbor) + 1);
                    } else {
                        map.put(neighbor, 1);
                    }
                }
            }
            
            for (DirectedGraphNode node : graph) {
                if (!map.containsKey(node)) {
                    result.add(node);
                }
            }
            
            int start = 0;
            while (start < result.size()) {
                DirectedGraphNode node = result.get(start++);
                for (DirectedGraphNode neighbor : node.neighbors) {
                    map.put(neighbor, map.get(neighbor) - 1);
                    if (map.get(neighbor) == 0) {
                        result.add(neighbor);
                    }
                }
            }
            
            return result;
        }
    View Code

    3 Route Between Two Nodes in Graph   bfs   1  hold visited  and queue  2 bfs

        public boolean hasRoute(ArrayList<DirectedGraphNode> graph, 
                                DirectedGraphNode s, DirectedGraphNode t) {
            if (s == t) {
                return true;
            }
            
            Queue<DirectedGraphNode> queue = new LinkedList<>();
            Set<DirectedGraphNode> visited = new HashSet<>();
            queue.add(s);
            visited.add(s);
            
            while (!queue.isEmpty()){
                DirectedGraphNode node = queue.poll();
                for (DirectedGraphNode neighbor : node.neighbors) {
                    if (visited.contains(neighbor)) {
                        continue;
                    }
                    queue.add(neighbor);
                    visited.add(neighbor);
                    if (neighbor == t) {
                        return true;
                    }
                }
            }
            
            return false;
        }
    View Code

    4 N-Queens

        public ArrayList<ArrayList<String>> solveNQueens(int n) 
        {
            // write your code here
            ArrayList<ArrayList<String>> res = new ArrayList<>();
            search(res, new ArrayList<Integer>(), n);
            return res;
        }
        void search(ArrayList<ArrayList<String>> res, ArrayList<Integer> cols, int n) {
            if (cols.size() == n) {
                res.add(drawChessboard(cols));
                return;
            }
            
            for (int i = 0; i < n; i++) {
                if (!isValid(cols, i)) {
                    continue;
                }
                cols.add(i);
                search(res, cols, n);
                cols.remove(cols.size() - 1);
            }
        }
        
        ArrayList<String> drawChessboard(List<Integer> cols) {
            ArrayList<String> result = new ArrayList<>();
            for (int i = 0; i < cols.size(); i++) {
                StringBuilder sb = new StringBuilder();
                for (int j = 0; j < cols.size(); j++) {
                    sb.append(cols.get(i) == j ? "Q" : ".");
                }
                result.add(sb.toString());
            }
            return result;
        }
        
        boolean isValid(List<Integer> cols, int colIndex) {
            int row = cols.size();
            for (int i = 0; i < cols.size(); i++) {
                if (cols.get(i) == colIndex) {
                    return false;
                }
                
                if (i + cols.get(i) == row + colIndex) {
                    return false;
                }
                
                if (i - cols.get(i) == row - colIndex) {
                    return false;
                }
            }
            return true;
        }
    View Code

    5 Word Ladder

       public int ladderLength(String start, String end, Set<String> dict) 
        {
            if (dict == null) {
                return 0;
            }
            
            if (start.equals(end)) {
                return 1;
            }
            
            dict.add(end);
            Set<String> visited = new HashSet<>();
            Queue<String> queue = new LinkedList<>();
            queue.add(start);
            int length = 1;
            
            while (!queue.isEmpty()) {
                length++;
                int size = queue.size();
                for (int i = 0; i < size; i++) {
                    String word = queue.poll();
                    for (String newWord: getNextWords(word, dict)) {
                        if (visited.contains(newWord)) {
                            continue;
                        }
                        
                        if (newWord.equals(end)) {
                            return length;
                        }
                        
                        visited.add(newWord);
                        queue.add(newWord);
                    }
                }
            }
            return 0;
        }
        
        List<String> getNextWords(String word,  Set<String> dict) {
            List<String> result = new ArrayList<>();
            for (char c = 'a'; c <= 'z'; c++) {
                for (int i = 0; i < word.length(); i++) {
                    if (c == word.charAt(i)) {
                        continue;
                    }
                    String newWord = replace(word, i, c);
                    if (dict.contains(newWord)) {
                        result.add(newWord);
                    }
                }
            }
            return result;
        }
        
        
        String replace(String word, int i, char c) {
            char[] arr = word.toCharArray();
            arr[i] = c;
            return new String(arr);
        }
    View Code

     6 Word Ladder

    public class Solution {
        /**
          * @param start, a string
          * @param end, a string
          * @param dict, a set of string
          * @return a list of lists of string
          */
        public List<List<String>> findLadders(String start, String end, Set<String> dict) {
            List<List<String>> result = new ArrayList<>();
            Map<String, Integer> distance = new HashMap<>();
            Map<String, List<String>> map = new HashMap<>();
            
            dict.add(start);
            dict.add(end);
            
            bfs(start, end, dict, distance, map);
            
            List<String> path = new ArrayList<>();
            
            dfs(start, end, result, distance, map, path);
            
            return result;
        }
        
        void dfs(String start, String cur, List<List<String>> result, Map<String, Integer> distance, 
                 Map<String, List<String>> map, List<String> path) {
            path.add(cur);
            if (cur.equals(start)) {
                Collections.reverse(path);
                result.add(new ArrayList<>(path));
                Collections.reverse(path);
            } else {
            for (String word : map.get(cur)) {
                if (distance.containsKey(word) && distance.get(word) + 1 == distance.get(cur)) {
                    dfs(start, word, result, distance, map, path);
                }
            }
            }
            path.remove(path.size() - 1);
        }
        
        void bfs(String start, String end, Set<String> dict, Map<String, Integer> distance, 
                 Map<String, List<String>> map) {
                 
            Queue<String> queue = new LinkedList<>();
            queue.offer(start);
            distance.put(start, 0);
            
            for(String word : dict) {
                map.put(word, new ArrayList<String>());
            }
            
            while (!queue.isEmpty()) {
                String word = queue.poll();
                for (String newWord: getNextWords(word, dict)) {
                    map.get(newWord).add(word);
                    if (!distance.containsKey(newWord)) {
                        distance.put(newWord, distance.get(word) + 1);
                        queue.offer(newWord);
                    }
                }
            }
        }
        
            List<String> getNextWords(String word,  Set<String> dict) {
            List<String> result = new ArrayList<>();
            for (char c = 'a'; c <= 'z'; c++) {
                for (int i = 0; i < word.length(); i++) {
                    if (c == word.charAt(i)) {
                        continue;
                    }
                    String newWord = replace(word, i, c);
                    if (dict.contains(newWord)) {
                        result.add(newWord);
                    }
                }
            }
            return result;
        }
        
        
        String replace(String word, int i, char c) {
            char[] arr = word.toCharArray();
            arr[i] = c;
            return new String(arr);
        }
    }
    View Code

    7 Palindrome Partitioning

         public ArrayList<ArrayList<String>> partition(String s) {
             ArrayList<ArrayList<String>> res = new ArrayList<>();
             ArrayList<String> path = new ArrayList<>();
             dfs(s, 0, path, res);
             return res;
         }
         void dfs(String s, int start, ArrayList<String> path, ArrayList<ArrayList<String>> res) {
             if (start == s.length()) {
                 res.add(new ArrayList<>(path));
                 return;
             }
             for (int i = start; i < s.length(); i++) {
                 if (isValid(s, start, i)){
                     path.add(s.substring(start, i + 1));
                     dfs(s, i + 1, path, res);
                     path.remove(path.size() - 1);
                 }
             }
         }
         boolean isValid(String s, int left, int right) {
             while (left < right) {
                 if (s.charAt(left++) != s.charAt(right--)) {
                     return false;
                 }
             }
             return true;
         }
    View Code
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  • 原文地址:https://www.cnblogs.com/whesuanfa/p/7439989.html
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