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:
    Python解释器
    js子节点children和childnodes的用法
    添加jar包需注意
    Class.forName("com.mysql.jdbc.driver");
    java集合类总结
    interface思考练习一
    java.lang.ClassNotFoundException: com.mysql.jdbc.Driver
    Struts2的配置文件中, <package>的作用,<action><result>重名?
    在Struts2的Action中获得request response session几种方法
    学习一直都是一个相见恨晚的过程,我希望我的相见恨晚不会太晚。
  • 原文地址:https://www.cnblogs.com/jcliBlogger/p/4550994.html
Copyright © 2011-2022 走看看