码迷,mamicode.com
首页 > 其他好文 > 详细

搜索类题目的高效实现

时间:2017-10-31 12:59:36      阅读:211      评论:0      收藏:0      [点我收藏+]

标签:open   ide   alt   letter   static   square   []   new   for   

BFS 类问题

1 Surrounded Regions

技术分享
  public void surroundedRegions(char[][] board)
    {
        int n = board.length;
        if (n == 0) {
            return;
        }
        int m = board[0].length;
        for (int i = 0; i < m; i++) {
            bfs(board, 0, i);
            bfs(board, n - 1, i);
        }
        for (int i = 0; i < n; i++) {
            bfs(board, i, 0);
            bfs(board, i, m - 1);
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (board[i][j] == O) {
                    board[i][j] = X;
                } else if (board[i][j] == #) {
                    board[i][j] = O;
                }
            }
        }
    }
    void bfs(char[][] board, int i, int j) {
        if (board[i][j] != O) {
            return;
        }
        int[] dx = {1, -1, 0, 0};
        int[] dy = {0, 0, -1, 1};
        Queue<Integer> qx = new LinkedList<>();
        Queue<Integer> qy = new LinkedList<>();
        qx.offer(i);
        qy.offer(j);
        while (!qx.isEmpty()) {
            int cx = qx.poll();
            int cy = qy.poll();
            board[cx][cy] = #;
            for (int k = 0; k < dx.length; k++) {
                int nx = cx + dx[k];
                int ny = cy + dy[k];
                if (nx >= 0 && nx < board.length && ny >= 0 && ny < board[0].length && board[nx][ny] == O) {
                    qx.offer(nx);
                    qy.offer(ny);
                }
            }
        } 
    }
View Code

2 Nearest Exit

技术分享

技术分享
public class MovingAvage {
    static final int INF = 2147483647;
    int n, m;
    public void wallsAndGates(int[][] rooms) {
        n = rooms.length;
        if (n == 0) {
            return;
        }
        m = rooms[0].length;
        int[] dx = {0, 0, -1, 1};
        int[] dy = {-1, 1, 0, 1};
        Queue<Integer> qx = new LinkedList<>();
        Queue<Integer> qy = new LinkedList<>();
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (rooms[i][j] == 0) {
                    qx.offer(i);
                    qy.offer(j);
                }
            }
        }
        while (!qx.isEmpty()) {
            int cx = qx.poll();
            int cy = qy.poll();
            for (int i = 0; i < dx.length; i++) {
                int nx = cx + dx[i];
                int ny = cy + dy[i];
                if (0 <= nx && nx < n && 0 <= ny && ny < m && rooms[nx][ny] == INF) {
                    qx.offer(nx);
                    qy.offer(ny);
                    rooms[nx][ny] = rooms[cx][cy] + 1;
                }
            }
        }
    }
View Code

3  17. Letter Combinations of a Phone Number

技术分享
    public List<String> letterCombinations(String digits) {
        String[] map = new String[] {"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        List<String> ans = new ArrayList<>();
        if (digits.length() == 0) {
            return ans;
        }
        dfs(digits, 0, "", ans, map);
        return ans;
    }
    void dfs(String digits, int index, String item, List<String> ans, String[] map) {
        if (index == digits.length()) {
            ans.add(item);
            return;
        }
        int d = digits.charAt(index) - 0;
        for (char c : map[d].toCharArray()) {
            dfs(digits, index + 1, item + c, ans, map);
        }
    }
View Code

4 Factorization

技术分享

技术分享
    public List<List<Integer>> getFactors(int n) {
        List<List<Integer>> res = new ArrayList<>();
        List<Integer> path = new ArrayList<>();
        dfs(2, n, res, path);
        return res;
    }
    void dfs(int start, int remain, List<List<Integer>> res, List<Integer> path) {
        if (remain == 1) {
            if (path.size() != 1) {
                res.add(new ArrayList<>(path));
            }
            return;
        }
        for (int i = start; i <= remain; i++) {
            if (i > remain / i) {
                break;
            }
            if (remain % i == 0) {
                path.add(i);
                dfs(start, remain / i, res, path);
                path.remove(path.size() - 1);
            }
        }
        path.add(remain);
        dfs(remain, 1);
        path.remove(path.size() - 1);
    }
View Code

5 Word Squares

技术分享

 

技术分享
public class MovingAvage {
    int wordLen;
    List<String> squares = new ArrayList<>();
    Map<String, List<String>> hash = new HashMap<>();
    List<List<String>> ans = new ArrayList<>();
    
    public List<List<String>> wordSquares(String[] words) {
        if (words.length == 0) {
            return ans;
        }
        initPrefix(words);
        wordLen = words[0].length();
        dfs(0);
        return ans;
    }

    private void dfs(int l) {
        // TODO Auto-generated method stub
        if (l == wordLen) {
            ans.add(new ArrayList<>(squares));
            return;
        }
        String pre = "";
        for (int i = 0; i < l; i++) {
            pre += squares.get(i).charAt(l);
        }
        List<String> w = hash.get(pre);
        for (String item : w) {
            if (!checkPrefix(l, item)) {
                continue;
            }
            squares.add(item);
            dfs(l + 1);
            squares.remove(squares.size() - 1);
        }
    }

    private boolean checkPrefix(int l, String item) {
        // TODO Auto-generated method stub
        for (int j = l + 1; j < wordLen; j++) {
            String pre = "";
            for (int k = 0; k < l; k++) {
                pre = pre + squares.get(k).charAt(j);
            }
            pre += item.charAt(j);
            if (hash.containsKey(pre)) {
                return false;
            }
        }
        return true;
    }

    private void initPrefix(String[] words) {
        // TODO Auto-generated method stub
        for (String item : words) {
            if (!hash.containsKey("")) {
                hash.put("", new ArrayList<String>());
            }
            hash.get("").add(item);
            String pre = "";
            for (char c : item.toCharArray()) {
                pre += c;
                if (!hash.containsKey(pre)) {
                    hash.put(pre, new ArrayList<String>());
                }
                hash.get("pre").add(item);
            }
          }
    }
}
View Code

 

6 Add Operators

技术分享
     String num;
     int target;
     List<String> res = new ArrayList<>();
    public List<String> addOperators(String num, int target) {
        // write your code here
        this.num = num;
        this.target = target;
        dfs(0, "", 0, 0);
        return res;
    }
    void dfs(int start, String item, long sum, long last) {
        if (start == num.length()) {
            if (sum == target) {
                res.add(item);
            }
            return;
        }
        for (int i = start; i < num.length(); i++) {
            long x = Long.parseLong(num.substring(start, i + 1));
            if (start == 0) {
                dfs(i + 1, "" + x, x, x);
            } else {
                dfs(i + 1, item + "+" + x, sum + x, x);
                dfs(i + 1, item + "-" + x, sum - x, -x);
                dfs(i + 1, item + "*" + x, sum - last + last * x, last * x);
            }
            if (x == 0) {
                break;
            }
        }
    }
View Code

 

搜索类题目的高效实现

标签:open   ide   alt   letter   static   square   []   new   for   

原文地址:http://www.cnblogs.com/whesuanfa/p/7760545.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!