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,"SEE", -> returns true,"ABCB", -> returns false.
Array Backtracking
class Solution { vector<vector<bool> > canUse; public: bool dfs(int dep, int x, int y, vector<vector<char> >& board, string& word) { if(dep == word.size()) return true; int row = board.size(); int col = board[0].size(); if(x < 0 || x >= row || y <0 || y >= col ) return false; if(board[x][y] != word[dep]) return false; //mark canUse[x][y] = false; if(x < row-1 && canUse[x+1][y] ) { if(dfs(dep+1, x+1, y, board, word)) { //mark canUse[x][y] = true; return true; } } if(x >0 && canUse[x-1][y] ) { if(dfs(dep+1, x-1, y, board, word)) { //mark canUse[x][y] = true; return true; } } if(y < col-1 && canUse[x][y+1]) { if(dfs(dep+1, x, y+1, board, word)) { //mark canUse[x][y] = true; return true; } } if(y >0 && canUse[x][y-1]) { if(dfs(dep+1, x, y-1, board, word)) { //mark canUse[x][y] = true; return true; } } canUse[x][y] = true; return false; } bool exist(vector<vector<char> > &board, string word) { int row = board.size(); if(row == 0) return false; int col = board[0].size(); if(word.size() > row*col) return false; canUse.clear(); vector<bool> tmp(col, true); canUse.resize(row, tmp); for(int i = 0; i< row; i++) { for(int j = 0; j< col; j++) { if(dfs(0, i, j, board, word )) return true; } } return false; } };
原文：http://www.cnblogs.com/diegodu/p/4343924.html