LeetCode //C - 79. Word Search

Given an m x n grid of characters board and a string word , return true if word exists in the grid.

The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.

Example 1:

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

Example 2:

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

Example 3:

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

Constraints:
  • m == board.length
  • n = board[i].length
  • 1 <= m, n <= 6
  • 1 <= word.length <= 15
  • board and word consists of only lowercase and uppercase English letters.

From: LeetCode

Link: 79. Word Search


Solution:

Ideas:

Main Idea:

The main concept here is to use Depth First Search (DFS) to explore the board. Starting from each cell, the algorithm tries to construct the word by moving either horizontally or vertically through adjacent cells. While doing so, the algorithm makes sure not to use the same cell more than once.

Details:
1. Starting Point:

  • The function exist iterates through every cell in the board. For each cell, it checks if the word can be formed starting from that cell by invoking the dfs function.

2. DFS Function (dfs):

  • The purpose of this function is to explore all possible paths from a given cell (i, j) to see if the word can be constructed.
  • It first checks if the current cell's value matches the current character of the word (word[index]). If not, it returns false.
  • If the cell's value matches the last character of the word, it means the word is found, and it returns true.
  • If the cell's value matches the current character but is not the last character of the word, it proceeds to search in all four directions: up, down, left, and right.

3. Avoiding Revisiting the Same Cell:

  • To ensure the same cell is not used more than once, the function temporarily marks the current cell as visited by setting its value to 0 (or any character that is not a valid board character). This is a kind of "backtracking".
  • After exploring all paths from the current cell, its original value is restored, which undoes the "marking".

4. Result:

  • If at any point during the search, the dfs function finds the word, it returns true to the main exist function.
  • If the word is not found starting from any cell in the board, the exist function returns false.
Code:
c 复制代码
bool dfs(char** board, int i, int j, char* word, int index, int boardSize, int boardColSize) {
    // Base case: if the current position is out of bounds or the current character does not match
    if (i < 0 || i >= boardSize || j < 0 || j >= boardColSize || board[i][j] != word[index]) {
        return false;
    }

    // If we've reached the end of the word, then we've found a match
    if (index == strlen(word) - 1) {
        return true;
    }

    char tmp = board[i][j];
    board[i][j] = 0;  // Mark the cell as visited

    // Recursively search for the next character in all four directions (up, down, left, right)
    bool found = dfs(board, i + 1, j, word, index + 1, boardSize, boardColSize)
              || dfs(board, i - 1, j, word, index + 1, boardSize, boardColSize)
              || dfs(board, i, j + 1, word, index + 1, boardSize, boardColSize)
              || dfs(board, i, j - 1, word, index + 1, boardSize, boardColSize);

    board[i][j] = tmp;  // Restore the cell value after the DFS

    return found;
}

bool exist(char** board, int boardSize, int* boardColSize, char* word) {
    for (int i = 0; i < boardSize; i++) {
        for (int j = 0; j < boardColSize[0]; j++) {
            // Start the DFS search from each cell in the board
            if (dfs(board, i, j, word, 0, boardSize, boardColSize[0])) {
                return true;
            }
        }
    }
    return false;
}
相关推荐
老马啸西风9 分钟前
力扣 LC27. 移除元素 remove-element
算法·面试·github
数智顾问13 分钟前
中秋特别篇:使用QtOpenGL和着色器绘制星空与满月——从基础框架到光影渲染
算法
txwtech16 分钟前
第5篇 如何计算两个坐标点距离--opencv图像中的两个点
人工智能·算法·机器学习
CoovallyAIHub16 分钟前
YOLO26学界首评:四大革新点究竟有多强?
深度学习·算法·计算机视觉
用户9163574409517 分钟前
LeetCode热题100——11.盛最多水的容器
javascript·算法
迎風吹頭髮19 分钟前
UNIX下C语言编程与实践59-UNIX TCP 数据传输:send 与 recv 函数的使用与数据处理
c语言·网络·unix
迎風吹頭髮23 分钟前
UNIX下C语言编程与实践55-TCP 协议基础:面向连接的可靠传输机制与三次握手、四次挥手
c语言·网络·unix
Gorgous—l30 分钟前
数据结构算法学习:LeetCode热题100-矩阵篇(矩阵置零、螺旋矩阵、旋转图像、搜索二维矩阵 II)
数据结构·学习·算法
2401_841495641 小时前
【计算机视觉】霍夫变换函数的参数调整
人工智能·python·算法·计算机视觉·霍夫变换·直线检测·调整策略
练习前端两年半1 小时前
🔍 你真的会二分查找吗?
前端·javascript·算法