【Leetcode】Word Search

题目链接：https://leetcode.com/problems/word-search/

题目：

Given a 2D board and a word, find if the word exists in the grid.

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

For example,
Given board =

[  ['A','B','C','E'],  ['S','F','C','S'],  ['A','D','E','E']]

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

思路：

迷宫问题，本来很简单的题却总是超时，然后看了别人的解才发现要注意剪枝，在到达终点时候就要停止递归了，如果继续往下走，会超时。

算法1：会超时

int c[] = { 0, 1, -1, 0, 0 }, d[] = { 0, 0, 0, 1, -1 };int[][] maps;boolean flag = false;char board[][];String word;StringBuffer s = new StringBuffer();public boolean exist(char[][] board, String word) {this.board = board;maps = new int[board.length][board[0].length];this.word = word;boolean isFind = false;for (int i = 0; i < board.length; i++) {for (int j = 0; j < board[0].length; j++) {if (board[i][j] == word.charAt(0)) {searchRecursion(i, j, 0);if(flag)return true;}}}return false;}public void searchRecursion(int x, int y, int cur) {if (cur == word.length()) {flag = true;return;}for (int i = 0; i < 5; i++) {if (x + c[i] >= 0 && x + c[i] < maps.length && y + d[i] >= 0 && y + d[i] < maps[0].length) {if (maps[x + c[i]][y + d[i]] == 0 && board[x + c[i]][y + d[i]] == word.charAt(cur)) {searchRecursion(x + c[i], y + d[i], cur + 1);maps[x + c[i]][y + d[i]] = 0;}}}}

算法2：

int c[] = { 0, 1, -1, 0, 0 }, d[] = { 0, 0, 0, 1, -1 };int[][] maps; // 表示是否走过char board[][];String word;StringBuffer s = new StringBuffer();public boolean exist(char[][] board, String word) {this.board = board;maps = new int[board.length][board[0].length];this.word = word;for (int i = 0; i < board.length; i++) {for (int j = 0; j < board[0].length; j++) {if (board[i][j] == word.charAt(0) && searchRecursion(i, j, 0)) {return true;}}}return false;}public boolean searchRecursion(int x, int y, int cur) {if (cur == word.length()) {return true;}for (int i = 0; i < 5; i++) {if (x + c[i] >= 0 && x + c[i] < maps.length && y + d[i] >= 0 && y + d[i] < maps[0].length) {if (maps[x + c[i]][y + d[i]] == 0 && board[x + c[i]][y + d[i]] == word.charAt(cur)) { // 0可走maps[x + c[i]][y + d[i]] = 1;if (searchRecursion(x + c[i], y + d[i], cur + 1)) {return true;}maps[x + c[i]][y + d[i]] = 0;}}}return false;}