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搜索類題目的高效實現

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BFS 類問題

1 Surrounded Regions

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  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); } } } }
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2 Nearest Exit

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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;
                }
            }
        }
    }
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3 17. Letter Combinations of a Phone Number

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    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);
        }
    }
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4 Factorization

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    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);
    }
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5 Word Squares

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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);
            }
          }
    }
}
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6 Add Operators

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     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;
            }
        }
    }
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搜索類題目的高效實現