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
.public class Solution { public boolean exist(char[][] board, String word) { if(board == null || board.length == 0 || board[0].length == 0) return false; boolean res = false; for(int i = 0; i < board.length; i++) { for(int j = 0; j < board[0].length; j++) { if(board[i][j] == word.charAt(0)) { char[][] tmp = board; tmp[i][j] = 0; if(ifFound(tmp, i, j, word, 1)) return true; } } } return res; } public boolean ifFound(char[][] b, int i, int j, String s, int idx) { boolean res = false; if(idx == s.length()) return true; char c = s.charAt(idx); // top if(i- 1 >= 0 && b[i - 1][j] == c) { if(idx == s.length() - 1) return true; char[][] tmp = b; tmp[i - 1][j] = 0; if(ifFound(tmp, i - 1, j, s, idx + 1)) return true; } // bottom if(i + 1 < b.length && b[i + 1][j] == c) { if(idx == s.length() - 1) return true; char[][] tmp = b; tmp[i + 1][j] = 0; if(ifFound(tmp, i + 1, j, s, idx + 1)) return true; } // left if(j - 1 >= 0 && b[i][j - 1] == c) { if(idx == s.length() - 1) return true; char[][] tmp = b; tmp[i][j - 1] = 0; if(ifFound(tmp, i, j - 1, s, idx + 1)) return true; } // right if(j + 1 < b[0].length && b[i][j + 1] == c) { if(idx == s.length() - 1) return true; char[][] tmp = b; tmp[i][j + 1] = 0; if(ifFound(tmp, i, j + 1, s, idx + 1)) return true; } return res; } }
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