【问题描述】[中等]

【解答思路】

1. DFS繁琐版本

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优化版本

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}};// 盘面上有多少行private int m;// 盘面上有多少列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
链接：https://leetcode-cn.com/problems/word-search/solution/zai-er-wei-ping-mian-shang-shi-yong-hui-su-fa-pyth/

【总结】

1. 一开始没审题 以后是不回头的遍历 审题!!!

2.往往空间优先于时间

3.回溯算法相关题目

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[Leetcode][第39题][JAVA][组合总和][回溯][dfs][剪枝]

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