# 79. Word Search 单词搜索

@TOC

## # 题目描述

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.
``````

## # 解题方法

### # 回溯法

``````class Solution(object):
def exist(self, board, word):
"""
:type board: List[List[str]]
:type word: str
:rtype: bool
"""
for y in xrange(len(board)):
for x in xrange(len(board[0])):
if self.exit(board, word, x, y, 0):
return True
return False

def exit(self, board, word, x, y, i):
if i == len(word):
return True
if x < 0 or x >= len(board[0]) or y < 0 or y >= len(board):
return False
if board[y][x] != word[i]:
return False
board[y][x] = board[y][x].swapcase()
isexit =  self.exit(board, word, x + 1, y, i + 1) or self.exit(board, word, x, y + 1, i + 1) or self.exit(board, word, x - 1, y, i + 1) or self.exit(board, word, x, y - 1, i + 1)
board[y][x] = board[y][x].swapcase()
return isexit
``````

``````class Solution {
public:
bool exist(vector<vector<char>>& board, string word) {
if (word.size() == 0) return false;
const int M = board.size(), N = board[0].size();
vector<vector<bool>> visited(M, vector<bool>(N, false));
for (int i = 0; i < M; i++) {
for (int j = 0; j < N; j ++) {
if (dfs(board, word, 0, {i, j}, visited)) {
return true;
}
}
}
return false;
}
bool dfs(const vector<vector<char>>& board, const string& word, int start, pair<int, int> curpos, vector<vector<bool>>& visited) {
const int M = board.size(), N = board[0].size();
if (start == word.size()) return true;
if (curpos.first < 0 || curpos.first >= M || curpos.second < 0 || curpos.second >= N || visited[curpos.first][curpos.second] || word[start] != board[curpos.first][curpos.second])
return false;
visited[curpos.first][curpos.second] = true;
for (auto d : dirs) {
int nx = curpos.first + d.first;
int ny = curpos.second + d.second;
if (dfs(board, word, start + 1, {nx, ny}, visited))
return true;
}
visited[curpos.first][curpos.second] = false;
return false;
}
private:
vector<pair<int, int>> dirs = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
};
``````

## # 日期

2018 年 2 月 27 日 2018 年 12 月 22 日 —— 今天冬至