zoukankan      html  css  js  c++  java
  • [LeetCode] Word Search

    This problem is somewhat tricky at first glance. However, the final implementation is fairly simple using recursion.

    The basic idea is, visiting every possible position (i, j) of  board and find if starting from (i, j), the word exists in the board. For the word to exist starting from (i, j), two conditions should be satisfied:

    1. board[i][j] = word[0] ;
    2. Define tail = word.substr(1, word.length() - 1) , then we should be able to find tail starting from one of the four neighbors of (i, j), which requires a recursive call of the function.

    Putting 1 and 2 together, we have the following code, which is self-explanatory.

     1 class Solution {
     2 public:
     3     bool exist(vector<vector<char>> board, string word) {
     4         for (int i = 0; i < board.size(); i++)
     5             for (int j = 0; j < board[0].size(); j++)
     6                 if (isExist(board, word, i, j)) return true;
     7         return false;
     8     }
     9 private:
    10     bool isExist(vector<vector<char>>& board, string& word, int row, int col) {
    11         if (row < 0 || row >= board.size() || col < 0 || col >= board[0].size() || board[row][col] != word[0])
    12             return false;
    13         if (word.length() == 1) return true;
    14         board[row][col] = ' ';
    15         string tail = word.substr(1, word.length() - 1);
    16         if (isExist(board, tail, row - 1, col) || isExist(board, tail, row + 1, col) ||
    17             isExist(board, tail, row, col - 1) || isExist(board, tail, row, col + 1))
    18             return true;
    19         board[row][col] = word[0];
    20         return false;
    21     }
    22 };

    Note that in the above code, in order to prevent visiting the same grid of board again, we modify the starting point of the grid and recover it again in lines 14 and 19 respectively.

    The above code is readable, though not fast enough. A simple trick to make it faster to pass a char*  instead of string to the isExist function.

     1 class Solution {
     2 public:
     3     bool exist(vector<vector<char>> board, string word) {
     4         for (int i = 0; i < board.size(); i++)
     5             for (int j = 0; j < board[0].size(); j++)
     6                 if (isExist(board, word.c_str(), i, j)) return true;
     7         return false;
     8     }
     9 private:
    10     bool isExist(vector<vector<char>>& board, const char* word, int row, int col) {
    11         if (row < 0 || row >= board.size() || col < 0 || col >= board[0].size() || board[row][col] != *word)
    12             return false;
    13         if (!(*(word + 1))) return true;
    14         board[row][col] = ' ';
    15         if (isExist(board, word + 1, row - 1, col) || isExist(board, word + 1, row + 1, col) ||
    16             isExist(board, word + 1, row, col - 1) || isExist(board, word + 1, row, col + 1))
    17             return true;
    18         board[row][col] = *word;
    19         return false;
    20     }
    21 };
  • 相关阅读:
    博客CSS样式 二
    产品经理
    HttpClient调用doGet、doPost、JSON传参及获得返回值
    Maven无法下载com.oracle:ojdbc.jar解决方法
    EasyExcel导入导出
    centos7 安装supervisor教程以及常见问题
    Django与Celery的安装使用
    亚马逊广告api v2版php扩展库
    Mac VMware Fusion CentOS7 安装、配置静态IP文档
    常见Web安全问题攻防解析
  • 原文地址:https://www.cnblogs.com/jcliBlogger/p/4550994.html
Copyright © 2011-2022 走看看