【LeetCode】Word Search
2014-06-07 14:26
204 查看
参考链接
http://blog.csdn.net/doc_sgl/article/details/13008175题目描述
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.
For example,
Given board =
[ ["ABCE"], ["SFCS"], ["ADEE"] ]
word =
"ABCCED", -> returns
true,
word =
"SEE", -> returns
true,
word =
"ABCB", -> returns
false.
题目分析
总结
代码示例
class Solution { public: bool exist(vector<vector<char> > &board, string word) { int m = board.size(); if(m < 1) return false; if(word.size() < 1) return false; int n = board[0].size(); vector<vector<bool> > used(m, vector<bool>(n,false)); for(int i = 0;i<m;i++) { for(int j = 0;j<n;j++) { if(board[i][j] == word[0] && existCore(board,word,i,j,0,used)) return true; } } return false; } bool existCore(vector<vector<char> > &board, string word,int i,int j,int index,vector<vector<bool> > &used) { if(index == word.size())/////////////// return true; if(i<0 || i>=board.size() || j <0 || j>=board[0].size()) return false; if(used[i][j] || board[i][j] != word[index]) return false; used[i][j] = true; if(existCore(board,word,i+1,j,index+1,used)) return true;///// if(existCore(board,word,i-1,j,index+1,used)) return true;////// if(existCore(board,word,i,j+1,index+1,used)) return true;/////// if(existCore(board,word,i,j-1,index+1,used)) return true;////////// used[i][j] = false; return false; } };
bool dfs(vector<vector<char> > &board, int x, int y, string word) { if(word.size() == 0)return true; bool flag = false; if(x-1>=0 && board[x-1][y] == word[0]) { board[x-1][y] = '#'; flag = dfs(board, x-1, y, word.substr(1)); board[x-1][y] = word[0]; } if(!flag && y-1>=0 && board[x][y-1] == word[0]) { board[x][y-1] = '#'; flag = dfs(board, x, y-1, word.substr(1)); board[x][y-1] = word[0]; } if(!flag && x+1<board.size() && board[x+1][y] == word[0]) { board[x+1][y] = '#'; flag = dfs(board, x+1, y, word.substr(1)); board[x+1][y] = word[0]; } if(!flag && y+1<board[0].size() && board[x][y+1] == word[0]) { board[x][y+1] = '#'; flag = dfs(board, x, y+1, word.substr(1)); board[x][y+1] = word[0]; } return flag; } bool exist(vector<vector<char> > &board, string word) { // Note: The Solution object is instantiated only once. if(word.size() < 1)return true; int row = board.size(); int col = board[0].size(); set<string> st; for(int i = 0; i < row; i++) for(int j = 0; j < col; j++) if(board[i][j] == word[0]) { board[i][j] = '#'; if(dfs(board,i,j,word.substr(1))) return true; board[i][j] = word[0]; } return false; }
推荐学习C++的资料
C++标准函数库http://download.csdn.net/detail/chinasnowwolf/7108919
在线C++API查询
http://www.cplusplus.com/
相关文章推荐
- [leetcode] 212.Word Search II
- Word Search | LeetCode
- Leetcode之Word Search 问题
- [Leetcode]Word Search
- LeetCode-79-Word Search 爆搜
- [LeetCode] Word Search
- LeetCode | Word Search
- 【Leetcode】Word Search
- leetcode笔记:Word Search
- [LeetCode]105. Word Search单词查找
- 【LeetCode-面试算法经典-Java实现】【079-Word Search(单词搜索)】
- leetcode: word search
- leetcode 79. Word Search DFS 单词搜索 + 深度优先遍历
- (java)leetcode-79:Word Search
- 算法分析与设计课程——LeetCode刷题之 Word Search
- [LeetCode]Word Search
- LeetCode:Word Search
- 【LeetCode】Word Search
- 【leetcode】Word Search
- leetcode79 Word Search