LeetCode（算法）- 79. 单词搜索

题目大意：略

解题思路：略

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AC 代码

Java

// 解决方案(1)
class Solution {
 
    // 使用 Map 超时
//    private Map<Integer, Boolean> usedMap = new HashMap<>();
 
    private boolean ok = false;
 
    private boolean[][] path;
 
    public boolean exist(char[][] board, String word) {
        path = new boolean[board.length][board[0].length];
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[i].length; j++) {
                if (ok) {
                    return true;
                }
//                usedMap.clear();
                // 不需要每次都初始化, 比较耗时, 因为回溯的时候会恢复到原始状态
//                path = new boolean[board.length][board[0].length];
                handle(board, i, j, word, 0);
            }
        }
        return ok;
    }
 
    private void handle(char[][] board, int i, int j, String word, int ptr) {
        if (ok || i >= board.length || i < 0 || j >= board[0].length || j < 0) {
            return;
        }
 
//        Integer key = i * 10 + j;
        if (path[i][j]) {
            return;
        }
 
        if (board[i][j] == word.charAt(ptr)) {
            if (ptr + 1 >= word.length()) {
                ok = true;
                return;
            }
            path[i][j] = true;
//            usedMap.put(key, true);
            handle(board, i + 1, j, word, ptr + 1);
            handle(board, i - 1, j, word, ptr + 1);
            handle(board, i, j + 1, word, ptr + 1);
            handle(board, i, j - 1, word, ptr + 1);
//            usedMap.remove(key);
            path[i][j] = false;
        }
    }
}
 
// 解决方案(2)
class Solution {
    public boolean exist(char[][] board, String word) {
        char[] words = word.toCharArray();
        for(int i = 0; i < board.length; i++) {
            for(int j = 0; j < board[0].length; j++) {
                if(dfs(board, words, i, j, 0)) return true;
            }
        }
        return false;
    }
    boolean dfs(char[][] board, char[] word, int i, int j, int k) {
        if(i >= board.length || i < 0 || j >= board[0].length || j < 0 || board[i][j] != word[k]) return false;
        if(k == word.length - 1) return true;
        board[i][j] = '\0';
        boolean res = dfs(board, word, i + 1, j, k + 1) || dfs(board, word, i - 1, j, k + 1) || 
                      dfs(board, word, i, j + 1, k + 1) || dfs(board, word, i , j - 1, k + 1);
        board[i][j] = word[k];
        return res;
    }
}

C++

class Solution {
public:
    bool exist(vector<vector<char>>& board, string word) {
        rows = board.size();
        cols = board[0].size();
        for(int i = 0; i < rows; i++) {
            for(int j = 0; j < cols; j++) {
                if(dfs(board, word, i, j, 0)) return true;
            }
        }
        return false;
    }
private:
    int rows, cols;
    bool dfs(vector<vector<char>>& board, string word, int i, int j, int k) {
        if(i >= rows || i < 0 || j >= cols || j < 0 || board[i][j] != word[k]) return false;
        if(k == word.size() - 1) return true;
        board[i][j] = '\0';
        bool res = dfs(board, word, i + 1, j, k + 1) || dfs(board, word, i - 1, j, k + 1) || 
                      dfs(board, word, i, j + 1, k + 1) || dfs(board, word, i , j - 1, k + 1);
        board[i][j] = word[k];
        return res;
    }
};

LeetCode（算法）- 79. 单词搜索

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