【Word Search】cpp
2015-05-29 11:17
459 查看
题目:
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.
For example,
Given board =
word =
word =
word =
代码:
tips:
好好领悟一下dfs吧。。。
1. 这道题在主程序中有一个循环,如果一旦发现word的起点,就从这个位置开始dfs,看能否返回结果。
2. dfs的过程跟模板一样。
shit
============================================
第二次过这道题,逻辑清晰了很多,修正了两个笔误bug,AC了。
第二次的代码,比第一次过的代码还优化了额外空间结构。O(1)额外空间。
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.
For example,
Given board =
[ ["ABCE"], ["SFCS"], ["ADEE"] ]
word =
"ABCCED", -> returns
true,
word =
"SEE", -> returns
true,
word =
"ABCB", -> returns
false.
代码:
class Solution { public: bool exist(vector<vector<char> >& board, string word) { vector<vector<bool> > visited; for ( size_t i = 0; i < board.size(); ++i ) { vector<bool> tmp(board[i].size(),false); visited.push_back(tmp); } for ( int i = 0; i < board.size(); ++i ) { for ( int j = 0; j < board[i].size(); ++j ) { if (Solution::dfs(board, visited, i, j, word, 0)) return true; } } return Solution::dfs(board, visited, 0, 0, word, 0); } static bool dfs( vector<vector<char> >& board, vector<vector<bool> >& visited, int i, int j, string& word, int curr ) { if ( curr==word.size() ) return true; if ( i<0 || i==board.size() || j<0 || j==board[i].size() ) return false; if ( visited[i][j] ) return false; if ( board[i][j]!=word[curr] ) return false; if ( board[i][j]==word[curr] ) { visited[i][j] = true; if ( Solution::dfs(board, visited, i, j+1, word, curr+1)) return true; if ( Solution::dfs(board, visited, i+1, j, word, curr+1)) return true; if ( Solution::dfs(board, visited, i, j-1, word, curr+1)) return true; if ( Solution::dfs(board, visited, i-1, j, word, curr+1)) return true; } visited[i][j] = false; return false; } };
tips:
好好领悟一下dfs吧。。。
1. 这道题在主程序中有一个循环,如果一旦发现word的起点,就从这个位置开始dfs,看能否返回结果。
2. dfs的过程跟模板一样。
shit
============================================
第二次过这道题,逻辑清晰了很多,修正了两个笔误bug,AC了。
class Solution { public: bool exist(vector<vector<char> >& board, string word) { for ( int i=0; i<board.size(); ++i ) { for ( int j=0; j<board[i].size(); ++j ) { if ( board[i][j]==word[0] ) { board[i][j] = '#'; if ( Solution::dfs(board, i, j, word.substr(1,word.size()-1)) ) return true; board[i][j] = word[0]; } } } return false; } static bool dfs( vector<vector<char> >& board, int i, int j, string word) { if ( word.size()==0 ) return true; // left if ( j-1>=0 && board[i][j-1]==word[0] ) { board[i][j-1] = '#'; if (Solution::dfs(board, i, j-1, word.substr(1,word.size()-1)) ) return true; board[i][j-1] = word[0]; } // right if ( j+1<board[0].size() && board[i][j+1]==word[0] ) { board[i][j+1] = '#'; if (Solution::dfs(board, i, j+1, word.substr(1,word.size()-1)) ) return true; board[i][j+1] = word[0]; } // up if ( i-1>=0 && board[i-1][j]==word[0] ) { board[i-1][j] = '#'; if (Solution::dfs(board, i-1, j, word.substr(1,word.size()-1)) ) return true; board[i-1][j] = word[0]; } // down if ( i+1<board.size() && board[i+1][j]==word[0] ) { board[i+1][j] = '#'; if (Solution::dfs(board, i+1, j, word.substr(1,word.size()-1)) ) return true; board[i+1][j] = word[0]; } return false; } };
第二次的代码,比第一次过的代码还优化了额外空间结构。O(1)额外空间。
相关文章推荐
- LeetCode的medium题集合(C++实现)十六
- OC与C++混编工程配置
- Android NDK学习 <五> C++ 支持
- C++ 8.1 面向对象的标准库(未完)
- c++中BOOL和bool的区别
- C++静态联编与动态联编
- 经典回溯算法——八后问题
- 趣味图形之 余弦函数cos与直线相交(另一种相交)
- C/C++之回调函数介绍
- 趣味图形之 余弦函数cos与直线相交
- 值得推荐的C/C++框架和库 (真的很强大)
- 趣味图形之 二次函数与圆相交
- C++指针--字符串与指针
- 趣味图形之 正弦函数sin&&余弦函数cos相交
- C++ 7.9 指向函数的指针
- C++学习 Boost状态机说明statechart
- C++读取系统当前时间
- C++读取系统当前时间 分类: C/C++ 2015-05-29 10:03 47人阅读 评论(0) 收藏
- c++中的继承、聚合、关联和组合
- [C语言] 对于scanf()函数使用%d格式说明符无法读取字符的个人理解