zoukankan      html  css  js  c++  java
  • [LC] 79. Word Search

    Given a 2D board and a word, find if the word exists in the grid.

    The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.

    Example:

    board =
    [
      ['A','B','C','E'],
      ['S','F','C','S'],
      ['A','D','E','E']
    ]
    
    Given word = "ABCCED", return true.
    Given word = "SEE", return true.
    Given word = "ABCB", return false.

    Time: O(M * N * 4^|Word|)
     1 class Solution {
     2     int row;
     3     int col;
     4     public boolean exist(char[][] board, String word) {
     5         row = board.length;
     6         col = board[0].length;
     7         boolean[][] visited = new boolean[row][col];
     8         for (int i = 0; i < row; i++) {
     9             for (int j = 0; j < col; j++) {
    10                 if (helper(board, visited, word, i, j, 0)) {
    11                     return true;
    12                 }
    13             }
    14         }
    15         return false;
    16     }
    17     
    18     private boolean helper(char[][] board, boolean[][] visited, String word, int i, int j, int index) {
    19         if (index == word.length()) {
    20             return true;
    21         }
    22         if (i < 0 || i >= row || j < 0 || j >= col) {
    23             return false;
    24         }
    25         if (word.charAt(index) == board[i][j] && !visited[i][j]) {
    26             visited[i][j] = true;
    27             boolean res = helper(board, visited, word, i + 1, j, index + 1) ||
    28                 helper(board, visited, word, i - 1, j, index + 1) ||
    29                 helper(board, visited, word, i, j + 1, index + 1) ||
    30                 helper(board, visited, word, i, j - 1, index + 1);
    31             // need to clean up visited
    32             visited[i][j] = false;
    33             return res;
    34         }
    35 
    36         return false;
    37     }
    38 }
    class Solution {
        int[][] directions = new int[][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        public boolean exist(char[][] board, String word) {
            boolean[][] visited = new boolean[board.length][board[0].length];
            for (int i = 0; i < board.length; i++) {
                for (int j = 0; j < board[0].length; j++) {
                    if (dfs(0, word, board, visited, i, j)) {
                        return true;   
                    }
                }
            }
            return false;
        }
        
        private boolean dfs(int index, String word, char[][] board, boolean[][] visited, int row, int col) {
            if (index == word.length()) {
                return true;
            }
            // need to check edge after base case e.g. [['A']] 'A', is actually out of board
            if (row < 0 || row >= board.length || col < 0 || col >= board[0].length || visited[row][col]) {
                    return false;
            }
            if (board[row][col] == word.charAt(index)) {
                visited[row][col] = true;
                for (int[] direction: directions) {
                    int nxtRow = direction[0] + row;
                    int nxtCol = direction[1] + col;
                    if (dfs(index + 1, word, board, visited, nxtRow, nxtCol)) {
                        return true;
                    }
                }
                visited[row][col] = false;
            }
            return false;
        }
    }
  • 相关阅读:
    js数组方法大全
    自定义函数实现10进制转化为16进制
    函数的递归
    自定义函数实现atoi功能
    自定义函数实现英文字母大小写的转化
    exit函数和return语句
    函数的返回值
    函数的形参与实参
    函数的定义与声明
    字符串的高级应用-char a[100] = "1+2=;3-2=;2*5=;8/4=;" 得到char a[100] ="1+2=3;3-2=1;2*5=10;8/4=2;"
  • 原文地址:https://www.cnblogs.com/xuanlu/p/12005363.html
Copyright © 2011-2022 走看看