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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.
思路：本题使用DFS求解即可。下面的解法假设输入word中不会包含‘#’这样的字符。
1 class Solution { 2 public: 3 bool exist( vector<vector<char>> &board, string word ) { 4 if( board.empty() ) { return word.empty(); } 5 rows = board.size(), cols = board[0].size(), slen = word.length(); 6 for( int i = 0; i < rows; ++i ) { 7 for( int j = 0; j < cols; ++j ) { 8 if( board[i][j] == word[0] && DFS( i, j, board, 0, word ) ) { return true; } 9 }10 }11 return false;12 }13 private:14 bool DFS( int i, int j, vector<vector<char>>& board, int index, const string& word ) {15 if( ++index == slen ) { return true; }16 board[i][j] = ‘#‘;17 if( i > 0 && board[i-1][j] == word[index] && DFS( i-1, j, board, index, word ) ) {18 board[i][j] = word[index-1];19 return true;20 }21 if( i < rows-1 && board[i+1][j] == word[index] && DFS( i+1, j, board, index, word ) ) {22 board[i][j] = word[index-1];23 return true;24 }25 if( j > 0 && board[i][j-1] == word[index] && DFS( i, j-1, board, index, word ) ) {26 board[i][j] = word[index-1];27 return true;28 }29 if( j < cols-1 && board[i][j+1] == word[index] && DFS( i, j+1, board, index, word ) ) {30 board[i][j] = word[index-1];31 return true;32 }33 board[i][j] = word[index-1];34 return false;35 }36 int rows, cols, slen;37 };
Word Search
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