79. 单词搜索:

给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中,返回 true ;否则,返回 false

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。

样例 1:

算法leetcode|79. 单词搜索(rust重拳出击)-LMLPHP

输入:
	
	board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
	
输出:
	
	true

样例 2:

算法leetcode|79. 单词搜索(rust重拳出击)-LMLPHP

输入:
	
	board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
	
输出:
	
	true

样例 3:

算法leetcode|79. 单词搜索(rust重拳出击)-LMLPHP

输入:

	board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
	
输出:
	
	false

提示:

  • m == board.length
  • n = board[i].length
  • 1 <= m, n <= 6
  • 1 <= word.length <= 15
  • board 和 word 仅由大小写英文字母组成

进阶:

  • 你可以使用搜索剪枝的技术来优化解决方案,使其在 board 更大的情况下可以更快解决问题?

分析:

  • 面对这道算法题目,二当家的再次陷入了沉思。
  • 需要尝试所有的可能,遍历是不可少的,使用循环或者递归,递归是首选,因为比较容易实现,深度优先,回溯,递归套娃大法好。
  • 需要遍历每一个元素,然后开始从每个位置使用递归套娃大法,分别向上,下,左,右四个方向进行递归,重复操作,直到匹配成功,或者匹配失败就回溯尝试其他,需要注意的是不走重复的位置,所以需要一个结构标记已经走过的位置,使用和原二维字符网格相同大小的结构比较容易,空间上应该还可以优化,但是没有深究,因为空间上并没有浪费,如果做优化,肯定是以牺牲效率为代价,所谓时间换空间,我不要换。

题解:

rust:

impl Solution {
    pub fn exist(board: Vec<Vec<char>>, word: String) -> bool {
        fn check(board: &Vec<Vec<char>>, word: &Vec<char>, idx: usize, visited: &mut Vec<Vec<bool>>, r: usize, c: usize) -> bool {
            if r >= visited.len() || c >= visited[0].len() || visited[r][c] || board[r][c] != word[idx] {
                return false;
            }
            if idx == word.len() - 1 {
                // 完全匹配
                return true;
            }
            visited[r][c] = true;
            // 上下左右递归套娃大法
            if check(board, word, idx + 1, visited, r - 1, c)
                || check(board, word, idx + 1, visited, r + 1, c)
                || check(board, word, idx + 1, visited, r, c - 1)
                || check(board, word, idx + 1, visited, r, c + 1)
            {
                return true;
            }
            visited[r][c] = false;
            return false;
        }

        let word = word.chars().collect::<Vec<_>>();
        let mut visited = vec![vec![false; board[0].len()]; board.len()];
        for r in 0..board.len() {
            for c in 0..board[0].len() {
                if check(&board, &word, 0, &mut visited, r, c) {
                    return true;
                }
            }
        }
        return false;
    }
}

go:

func exist(board [][]byte, word string) bool {
    visited := make([][]bool, len(board))
	for i := range visited {
		visited[i] = make([]bool, len(board[0]))
	}

	var check func(idx, r, c int) bool
	check = func(idx, r, c int) bool {
		if r < 0 || r >= len(board) || c < 0 || c >= len(board[0]) || visited[r][c] || board[r][c] != word[idx] {
			return false
		}
		if idx == len(word)-1 {
			// 完全匹配
			return true
		}
		visited[r][c] = true
		// 上下左右递归套娃大法
		if check(idx+1, r-1, c) || check(idx+1, r+1, c) || check(idx+1, r, c-1) || check(idx+1, r, c+1) {
			return true
		}
		visited[r][c] = false
		return false
	}

	for r, row := range board {
		for c := range row {
			if check(0, r, c) {
				return true
			}
		}
	}

	return false
}

c++:

class Solution {
private:
    bool check(vector<vector<char>>& board, string& word, int idx, vector<vector<bool>>& visited, int r, int c) {
        if (r < 0 || r >= visited.size() || c < 0 || c >= visited[0].size() || visited[r][c] || board[r][c] != word[idx]) {
            return false;
        }
        if (idx == word.size() - 1) {
            // 完全匹配
            return true;
        }
        visited[r][c] = true;
        // 上下左右递归套娃大法
        if (check(board, word, idx + 1, visited, r - 1, c)
            || check(board, word, idx + 1, visited, r + 1, c)
            || check(board, word, idx + 1, visited, r, c - 1)
            || check(board, word, idx + 1, visited, r, c + 1)) {
            return true;
        }
        visited[r][c] = false;
        return false;
    }
public:
    bool exist(vector<vector<char>>& board, string word) {
        vector<vector<bool>> visited(board.size(), vector<bool>(board[0].size(), false));

        for (int r = 0; r < board.size(); ++r) {
            for (int c = 0; c < board[0].size(); ++c) {
                if (check(board, word, 0, visited, r, c)) {
                    return true;
                }
            }
        }

        return false;
    }
};

python:

class Solution:
    def exist(self, board: List[List[str]], word: str) -> bool:
        def check(idx: int, r: int, c: int) -> bool:
            if r < 0 or r >= len(board) or c < 0 or c >= len(board[0]) or visited[r][c] or board[r][c] != word[idx]:
                return False
            if idx == len(word) - 1:
                return True
            visited[r][c] = True
            if check(idx + 1, r - 1, c) or check(idx + 1, r + 1, c) or check(idx + 1, r, c - 1) or check(idx + 1, r, c + 1):
                return True
            visited[r][c] = False
            return False

        visited = [[False] * len(board[0]) for _ in range(len(board))]
        for r in range(len(board)):
            for c in range(len(board[0])):
                if check(0, r, c):
                    return True

        return False


java:

class Solution {
    public boolean exist(char[][] board, String word) {
        boolean[][] visited = new boolean[board.length][board[0].length];

        for (int r = 0; r < board.length; ++r) {
            for (int c = 0; c < board[0].length; ++c) {
                if (check(board, word, 0, visited, r, c)) {
                    return true;
                }
            }
        }

        return false;
    }

    private boolean check(char[][] board, String word, int idx, boolean[][] visited, int r, int c) {
        if (r < 0 || r >= visited.length || c < 0 || c >= visited[0].length || visited[r][c] || board[r][c] != word.charAt(idx)) {
            return false;
        }
        if (idx == word.length() - 1) {
            // 完全匹配
            return true;
        }
        visited[r][c] = true;
        // 上下左右递归套娃大法
        if (check(board, word, idx + 1, visited, r - 1, c)
                || check(board, word, idx + 1, visited, r + 1, c)
                || check(board, word, idx + 1, visited, r, c - 1)
                || check(board, word, idx + 1, visited, r, c + 1)) {
            return true;
        }
        visited[r][c] = false;
        return false;
    }
}


09-07 17:16