[LeetCode] 79. 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 =
[
[‘A’,’B’,’C’,’E’],
[‘S’,’F’,’C’,’S’],
[‘A’,’D’,’E’,’E’]
]
word = “ABCCED”, -> returns true,
word = “SEE”, -> returns true,
word = “ABCB”, -> returns false.
在board中尋找word是否存在,board構造word的方法是可以上下左右移動去構造word。
直接模擬搜索方法。
class Solution {
public:
bool find_ans(vector<vector<char>>& board, vector<vector<bool>>& isVisit, string word, int i, int j, int k, int ilen, int jlen) {
if (board[i][j] != word[k]) {
return false;
}
if (k == word.length()-1) {
return true;
}
bool ans = false;
isVisit[i][j] = true;
if (i+1 < ilen) {
if (!isVisit[i+1][j]) ans = find_ans(board, isVisit, word, i+1, j, k+1, ilen, jlen);
if (ans) return true;
}
if (j+1 < jlen) {
if (!isVisit[i][j+1]) ans = find_ans(board, isVisit, word, i, j+1, k+1, ilen, jlen);
if (ans) return true;
}
if (i-1 >= 0) {
if (!isVisit[i-1][j]) ans = find_ans(board, isVisit, word, i-1, j, k+1, ilen, jlen);
if (ans) return true;
}
if (j-1 >= 0) {
if (!isVisit[i][j-1]) ans = find_ans(board, isVisit, word, i, j-1, k+1, ilen, jlen);
if (ans) return true;
}
isVisit[i][j] = false;
return false;
}
bool exist(vector<vector<char>>& board, string word) {
int ilen = board.size();
if (ilen == 0) {
return false;
}
int jlen = board[0].size();
if (jlen == 0) {
return false;
}
vector<vector<bool>> isVisit(ilen, vector<bool>(jlen, false));
for (int i=0; i<ilen; ++i) {
for (int j=0; j<jlen; ++j) {
if (find_ans(board, isVisit, word, i, j, 0, ilen, jlen)) {
return true;
}
}
}
return false;
}
};