212. Word Search II

原网址为https://leetcode.com/problems/word-search-ii/description/

一、题目

Given a 2D board and a list of words from the dictionary, find all words in the board.

Each word must 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 in a word.

For example,

Given words = 

["oath","pea","eat","rain"]

 and board =

[
['o','a','a','n'],
['e','t','a','e'],
['i','h','k','r'],
['i','f','l','v']
]

Return 

["eat","oath"]

.

Note:

You may assume that all inputs are consist of lowercase letters 

a-z

.

click to show hint.

二、代码

class Solution {
public:
vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
vector<string> ans;
sort(words.begin(), words.end());
for (int i = 0; i < words.size(); i++) {
if ((i == 0 || words[i] != words[i - 1]) && findWord(board, words[i]))
ans.push_back(words[i]);
}
return ans;
}

bool findWord(vector<vector<char>>& board, string& s) {
int row = board.size();
int col = board[0].size();

for (int i = 0; i < row; i++) {
for (int j = 0; j < col; j++) {
if (findWord(board, i, j, s, 0))
return true;
}
}

return false;
}

bool findWord(vector<vector<char>>& board, int i, int j, string& s, int k) {
int row = board.size();
int col = board[0].size();
if (k == s.size())
return true;
if (i < 0 || i >= row || j < 0 || j >= col)
return false;
if (board[i][j] == '1')
return false;
char tmp = board[i][j];
if (s[k] == board[i][j])
board[i][j] = '1';
else
return false;

bool flag = findWord(board, i - 1, j, s, k+1)
|| findWord(board, i + 1, j, s, k+1)
|| findWord(board, i, j - 1, s, k+1)
|| findWord(board, i, j + 1, s, k+1);
board[i][j] = tmp;
return flag;
}
};

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