给定一个二维网格和一个单词,找出该单词是否存在于网格中。
【LeetCode|LeetCode第79题--单词搜索】单词必须按照字母顺序,通过相邻的单元格内的字母构成,其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。
示例:
board =
[
[‘A’,‘B’,‘C’,‘E’],
[‘S’,‘F’,‘C’,‘S’],
[‘A’,‘D’,‘E’,‘E’]
]
给定 word = “ABCCED”, 返回 true.
给定 word = “SEE”, 返回 true.
给定 word = “ABCB”, 返回 false.
class Solution {
bool dfs(vector>& board, vector> &useLetter, string word, int index,
int row, int col)
{
int rowSize = board.size();
int colSize = board[0].size();
if(index >= word.length()) return true;
if(row-1 >= 0 && useLetter[row-1][col] == false && word[index] == board[row-1][col])
{
useLetter[row-1][col] = true;
if(dfs(board, useLetter, word, index+1, row-1, col)) return true;
useLetter[row-1][col] = false;
}
if(row+1 < rowSize && useLetter[row+1][col] == false && word[index] == board[row+1][col])
{
useLetter[row+1][col] = true;
//注意此处不能直接return dfs(board, useLetter, word, index+1, row+1, col);
if(dfs(board, useLetter, word, index+1, row+1, col)) return true;
useLetter[row+1][col] = false;
}
if(col-1 >= 0 && useLetter[row][col-1] == false && word[index] == board[row][col-1])
{
useLetter[row][col-1] = true;
if(dfs(board, useLetter, word, index+1, row, col-1)) return true;
useLetter[row][col-1] = false;
}
if(col+1 < colSize && useLetter[row][col+1] == false && word[index] == board[row][col+1])
{
useLetter[row][col+1] = true;
if(dfs(board, useLetter, word, index+1, row, col+1)) return true;
useLetter[row][col+1] = false;
}
return false;
}public:
bool exist(vector>& board, string word) {
if(board.size() == 0 || board[0].size() == 0 || word == "") return false;
int rowSize = board.size();
int colSize = board[0].size();
vector> useLetter(rowSize, vector(colSize, false));
for(int i = 0;
i < rowSize;
i++)
{
for(int j = 0;
j < colSize;
j++)
{
if(board[i][j] == word[0])//先找到第一个匹配的字母。
{
useLetter[i][j] = true;
if(dfs(board, useLetter, word, 1, i, j)) return true;
useLetter[i][j] = false;
}
}
}
return false;
}
};
理解深度优先遍历之后,该题并不难,原理同“解数独”类似,使用一个useLetter判断该字母是否被找到。
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