每日N+1题 -- 算法练习 Day3

题目主要来源：LeetCode。主要选自剑指offer、程序员面试宝典。
每日尽可能保持 N+1 道题，N 取 0 到 9。

部分解法是从LeetCode上大佬们的解法中拿过来的，如有侵权，告知立删

0x01 剑指 Offer 12. 矩阵中的路径

/**
 * JZ12  回溯
 *
 * @author Anjude
 * @date 2021/7/4 17:34
 */
public class JZ12 {
    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) {
        // 越界或者字符不匹配 -- 此路不通
        if (i >= board.length || i < 0 || j >= board[0].length || j < 0 || board[i][j] != word[k]) return false;
        // 最后一个字符匹配，返回true
        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;
    }

    public static void main(String[] args) {
        JZ12 jz12 = new JZ12();
        jz12.exist(new char[][]{{'a','b'},{'a','b'}}, "abb");
    }
}

0x02 剑指 Offer 05. 替换空格

/**
 * JZ5  可以用replace，也可以用StringBuilder遍历
 *
 * @author Anjude
 * @date 2021/7/4 22:05
 */
public class JZ5 {
    public String replaceSpace(String s) {
        StringBuilder sb = new StringBuilder();
        for(char c : s.toCharArray()){
            if(c == ' ') sb.append("%20");
            else sb.append(c);
        }
        return sb.toString();
    }
}

0x03 剑指 Offer 13. 机器人的运动范围

/**
 * JZ13  dfs
 *
 * @author Anjude
 * @date 2021/7/4 22:33
 */
public class JZ13 {
    public int movingCount(int m, int n, int k) {
        boolean[][] visited = new boolean[m][n];
        return dfs(0, 0, m, n, k, visited);
    }

    private int dfs(int i, int j, int m, int n, int k, boolean[][] visited) {
        // 终止条件：边界、不符合条件、已访问点
        if (i < 0 || i >= m || j < 0 || j >= n || (i / 10 + i % 10 + j / 10 + j % 10) > k || visited[i][j]) {
            return 0;
        }
        visited[i][j] = true;
        return dfs(i + 1, j, m, n, k, visited) + dfs(i - 1, j, m, n, k, visited) +
                dfs(i, j + 1, m, n, k, visited) + dfs(i, j - 1, m, n, k, visited) + 1;
    }
}