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
艾莉丝努力练剑21 分钟前
【LeetCode&数据结构】单链表的应用——反转链表问题、链表的中间节点问题详解
c语言·开发语言·数据结构·学习·算法·leetcode·链表
_殊途2 小时前
《Java HashMap底层原理全解析(源码+性能+面试)》
java·数据结构·算法
珊瑚里的鱼5 小时前
LeetCode 692题解 | 前K个高频单词
开发语言·c++·算法·leetcode·职场和发展·学习方法
不知道叫什么呀6 小时前
【C】vector和array的区别
java·c语言·开发语言·aigc
秋说6 小时前
【PTA数据结构 | C语言版】顺序队列的3个操作
c语言·数据结构·算法
lifallen7 小时前
Kafka 时间轮深度解析:如何O(1)处理定时任务
java·数据结构·分布式·后端·算法·kafka
liupenglove7 小时前
自动驾驶数据仓库:时间片合并算法。
大数据·数据仓库·算法·elasticsearch·自动驾驶
python_tty8 小时前
排序算法(二):插入排序
算法·排序算法
然我8 小时前
面试官:如何判断元素是否出现过?我:三种哈希方法任你选
前端·javascript·算法
F_D_Z9 小时前
【EM算法】三硬币模型
算法·机器学习·概率论·em算法·极大似然估计