【Leetcode 346/700】79. 單詞搜尋 【中等】【回溯深度搜尋JavaScript版】

叫我詹躲躲發表於2022-05-14

【Leetcode 346/700】79. 單詞搜尋 【中等】 回溯深度搜尋JavaScript版

1.題目

n 二維字元網格 board 和一個字串單詞 word 。如果 word 存在於網格中,返回 true ;否則,返回 false 。

單詞必須按照字母順序,通過相鄰的單元格內的字母構成,其中“相鄰”單元格是那些水平相鄰或垂直相鄰的單元格。同一個單元格內的字母不允許被重複使用。

示例 1:

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

示例 2:

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

示例 3:

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

提示:

    m == board.length
    n = board[i].length
    1 <= m, n <= 6
    1 <= word.length <= 15
    board 和 word 僅由大小寫英文字母組成

2解題思路

1.遍歷 board 所有元素,找到 word的第一個相同的元素,並且進行標記 (marked),進入遞迴去找接下來的第二個字元,接著第三個字母。如果沒找到,返回 false;

  1. 在設定的邊界內進行回溯搜尋,即上下左右進行搜尋下一個字元。找到了進入新的遞迴,沒有找到的話,直接返回false;

3.解題注意點

1.及時進行標記字元的狀態,是已經訪問了,還是未訪問;
2.如果最後所有的字串擷取完了,說明已經找到符合的答案啦,直接返回true;

4.解題程式碼

/**
 * @param {character[][]} board
 * @param {string} word
 * @return {boolean}
 */
var exist = function (board, word) {
    let border = [[0, 1], [0, -1], [1, 0], [-1, 0]], //定義上下左右四個方向
        col = board.length, //行數
        row = board[0].length, //列數
        marked = [...Array(col)].map(v => Array(row).fill()); //同行列空矩陣,用於記錄已經訪問的

        //空陣列直接返回false
    if (!col) return false;

    let backTracing = (i, j, markeds, boards, words) => {
        //擷取所有的字元,說明已經找到
        if (!words.length) {
            return true;
        }
        for (let p = 0; p < border.length; p++) {
            let curi = i + border[p][0]; //左右方向
            let curj = j + border[p][1]; //上下方向

            //判斷邊界,且找到了第一個字元
            if ((curi >= 0 && curi < col) && (curj >= 0 && curj < row && boards[curi][curj] == words[0])) {
                //已經用過,直接跳過
                if (markeds[curi][curj] == 1) {
                    continue
                }
                //標記為已使用
                markeds[curi][curj] = 1;
                //接著找下一個字元
                if (backTracing(curi, curj, markeds, boards, words.substring(1))) {
                    return true
                } else {
                    //使用完重置掉
                    markeds[curi][curj] = 0;
                }
            }
        }
        return false
    }

    for (let i = 0; i < col; i++) {
        for (let j = 0; j < row; j++) {
            if (board[i][j] === word[0]) {
                //找到第一個字元,標記為已經使用
                marked[i][j] = 1;
                //進入回溯
                if (backTracing(i, j, marked, board, word.substring(1))) {
                    return true
                } else {
                    //重置狀態
                    marked[i][j] = 0;
                }
            }
        }
    }
    return false
};
//測試用例 1
let board = [["A", "B", "C", "E"], ["S", "F", "C", "S"], ["A", "D", "E", "E"]], word = "ABCCED";

//測試用例2

let board1 = [["A", "B", "C", "E"], ["S", "F", "C", "S"], ["A", "D", "E", "E"]], word1 = "SEE"
//測試用例3
let board2 = [["A", "B", "C", "E"], ["S", "F", "C", "S"], ["A", "D", "E", "E"]], word2 = "ABCB"

console.log(exist(board, word))  //true
console.log(exist(board1, word1)) //true
console.log(exist(board2, word2)) //false

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