LeetCode: Word Search
2012-10-04 21:01
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Problem:
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 =
-> returns
word =
-> returns
word =
-> returns
递归搜索
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.
递归搜索
class Solution { public: bool find(vector< vector<char> > &board, int i, int j, string &word, int idx) { if (idx == word.size()) return true; if (i < board.size() && j < board[i].size() && i >= 0 && j >= 0 && board[i][j] == word[idx]) { board[i][j] = '#'; ++idx; return find(board, i+1, j, word, idx) | find(board, i-1, j, word, idx) | find(board, i, j-1, word, idx) | find(board, i, j+1, word, idx); } return false; } bool exist(vector<vector<char> > &board, string word) { // Start typing your C/C++ solution below // DO NOT write int main() function for (int i = 0; i < board.size(); ++i) { for (int j = 0; j < board[i].size(); ++j) { if (board[i][j] == word[0]) { vector<vector<char> > tmp (board); bool result = find(tmp, i, j, word, 0); if (result == true) return true; } } } } };
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