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:
    nmap扫描工具
    cobbler全自动批量安装部署linux
    使用ngxtop实时监控nginx
    Nginx 错误汇总
    定制sudo的密码保持时间以及如何不需要密码
    解决eclipse中出现Resource is out of sync with the file system问题
    log4j:ERROR Category option " 1 " not a decimal integer.错误解决
    properties文件中中文不能显示或者中文乱码
    Log4j使用总结
    eclipse, Log4j配置(真心的详细~)
  • 原文地址:https://www.cnblogs.com/jcliBlogger/p/4550994.html
Copyright © 2011-2022 走看看