【問題描述】[中等]

【解答思路】

1. DFS繁瑣版本

class Solution {public boolean exist(char[][] board, String word) {boolean flag = false;int row= board.length;int col= board[0].length;boolean[][] used = new boolean[row][col];for (int i = 0; i <row ; i++) {for (int j = 0; j <col ; j++) {flag = dfs(board,word,used,i,j,0);if(flag){return true;}}}return false;}private boolean dfs(char[][] board, String word,boolean[][] used,int x,int y ,int start) {int[] dx ={0,1,0,-1};int[] dy ={1,0,-1,0};boolean flag = false;int row= board.length;int col= board[0].length;if (start == word.length() - 1) {return board[x][y] == word.charAt(start);}if(board[x][y] == word.charAt(start)){used[x][y] =true;for (int i = 0; i <4 ; i++) {int newx = x+dx[i];int newy = y+dy[i];if(inArea(newx,newy,row,col) && !used[newx][newy]){flag = dfs(board,word,used,newx,newy,start+1);if(flag){return true;}}}used[x][y] =false;}return false;}private boolean inArea(int x,int y,int row,int col){return x>=0 && y>=0 && x<row && y<col;} }

2. DFS優化版本

public class Solution {private boolean[][] marked;// x-1,y// x,y-1 x,y x,y+1// x+1,yprivate int[][] direction = {{-1, 0}, {0, -1}, {0, 1}, {1, 0}};// 盤面上有多少行private int m;// 盤面上有多少列private int n;private String word;private char[][] board;public boolean exist(char[][] board, String word) {m = board.length;if (m == 0) {return false;}n = board[0].length;marked = new boolean[m][n];this.word = word;this.board = board;for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {if (dfs(i, j, 0)) {return true;}}}return false;}private boolean dfs(int i, int j, int start) {if (start == word.length() - 1) {return board[i][j] == word.charAt(start);}if (board[i][j] == word.charAt(start)) {marked[i][j] = true;for (int k = 0; k < 4; k++) {int newX = i + direction[k][0];int newY = j + direction[k][1];if (inArea(newX, newY) && !marked[newX][newY]) {if (dfs(newX, newY, start + 1)) {return true;}}}marked[i][j] = false;}return false;}private boolean inArea(int x, int y) {return x >= 0 && x < m && y >= 0 && y < n;}public static void main(String[] args) {// char[][] board = // { // {'A', 'B', 'C', 'E'}, // {'S', 'F', 'C', 'S'}, // {'A', 'D', 'E', 'E'} // }; // // String word = "ABCCED";char[][] board = {{'a', 'b'}};String word = "ba";Solution solution = new Solution();boolean exist = solution.exist(board, word);System.out.println(exist);} }作者：liweiwei1419 鏈接：https://leetcode-cn.com/problems/word-search/solution/zai-er-wei-ping-mian-shang-shi-yong-hui-su-fa-pyth/

【總結】

1. 一開始沒審題 以后是不回頭的遍歷 審題!!!

2.往往空間優先于時間

3.回溯算法相關題目

[Leedcode][JAVA][第46題][全排列][回溯算法]
[Leetcode][第81題][JAVA][N皇后問題][回溯算法]
[Leetcode][第60題][JAVA][第k個排列][回溯][DFS][剪枝]
[Leetcode][第39題][JAVA][組合總和][回溯][dfs][剪枝]

轉載鏈接：https://leetcode-cn.com/problems/word-search/solution/zai-er-wei-ping-mian-shang-shi-yong-hui-su-fa-pyth/

總結

以上是生活随笔為你收集整理的[Leetcode][第79题][JAVA][单词搜索][DFS][回溯]的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。