Word Search
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 =

[
  ["ABCE"],
  ["SFCS"],
  ["ADEE"]
]
word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

Solution1:

这个题目基本上就是DFS喽。然后注意一下递归回溯的问题。我们可以建设一个boolean的数组来记录访问过的值。在return false之前,我们应该把之前置

过的标记位置回来. 时间复杂度: http://www1.mitbbs.ca/article_t1/JobHunting/32524193_32524299_2.html

time 复杂度是m*n*4^(k-1). 也就是m*n*4^k.
m X n is board size, k is word size.

recuision最深是k层,recursive部分空间复杂度应该是O(k) + O(m*n)(visit array)

 package Algorithms.dfs;

 public class Exist {
public boolean exist(char[][] board, String word) {
if (board == null || word == null
|| board.length == 0 || board[0].length == 0) {
return false;
} int rows = board.length;
int cols = board[0].length; boolean[][] visit = new boolean[rows][cols]; // i means the index.
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
// dfs all the characters in the matrix
if (dfs(board, i, j, word, 0, visit)) {
return true;
}
}
} return false;
} public boolean dfs(char[][] board, int i, int j, String word, int wordIndex, boolean[][] visit) {
int rows = board.length;
int cols = board[0].length; int len = word.length();
if (wordIndex >= len) {
return true;
} // the index is out of bound.
if (i < 0 || i >= rows || j < 0 || j >= cols) {
return false;
} // the character is wrong.
if (word.charAt(wordIndex) != board[i][j]) {
return false;
} // 不要访问访问过的节点
if (visit[i][j] == true) {
return false;
} visit[i][j] = true; // 递归
// move down
if (dfs(board, i + 1, j, word, wordIndex + 1, visit)) {
return true;
} // move up
if (dfs(board, i - 1, j, word, wordIndex + 1, visit)) {
return true;
} // move right
if (dfs(board, i, j + 1, word, wordIndex + 1, visit)) {
return true;
} // move left
if (dfs(board, i, j - 1, word, wordIndex + 1, visit)) {
return true;
} // 回溯
visit[i][j] = false;
return false;
}
}

Solution2:

与解1是一样的,但我们可以省掉O(m*n)的空间复杂度。具体的作法是:在进入DFS后,把访问过的节点置为'#',访问完毕之后再置回来即可。

 /*
* Solution 2: 可以省掉一个visit的数组
* */
public boolean exist(char[][] board, String word) {
if (board == null || word == null
|| board.length == 0 || board[0].length == 0) {
return false;
} int rows = board.length;
int cols = board[0].length; // i means the index.
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
// dfs all the characters in the matrix
if (dfs(board, i, j, word, 0)) {
return true;
}
}
} return false;
} public boolean dfs(char[][] board, int i, int j, String word, int wordIndex) {
int rows = board.length;
int cols = board[0].length; int len = word.length();
if (wordIndex >= len) {
return true;
} // the index is out of bound.
if (i < 0 || i >= rows || j < 0 || j >= cols) {
return false;
} // the character is wrong.
if (word.charAt(wordIndex) != board[i][j]) {
return false;
} // mark it to be '#', so we will not revisit this.
board[i][j] = '#'; // 递归
// move down
if (dfs(board, i + 1, j, word, wordIndex + 1)) {
return true;
} // move up
if (dfs(board, i - 1, j, word, wordIndex + 1)) {
return true;
} // move right
if (dfs(board, i, j + 1, word, wordIndex + 1)) {
return true;
} // move left
if (dfs(board, i, j - 1, word, wordIndex + 1)) {
return true;
} // 回溯
board[i][j] = word.charAt(wordIndex);
return false;
}

December 16th, 2014 重写,简洁一点点,11分钟1次AC.

其实这里的回溯有一个点要注意,就是当dfs返回true时,我们并没有回溯。原因是这时整个dfs就结束了,我们也就不需要再回溯。否则一般的递归/回溯结构

这里是需要先回溯再返回的。

 public class Solution {
public boolean exist(char[][] board, String word) {
if (board == null || board.length == 0 || board[0].length == 0) {
return false;
} int rows = board.length;
int cols = board[0].length;
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
// Check if there is a word begin from i,j.
if (dfs(board, word, i, j, 0)) {
return true;
}
}
} return false;
} public boolean dfs(char[][] board, String word, int i, int j, int index) {
int len = word.length();
if (index >= len) {
return true;
} // Check the input parameter.
if (i < 0 || i >= board.length || j < 0 || j >= board[0].length) {
return false;
} // If the current character is right.
if (word.charAt(index) != board[i][j]) {
return false;
} char tmp = board[i][j];
board[i][j] = '#'; // recursion.
if (dfs(board, word, i + 1, j, index + 1)
|| dfs(board, word, i - 1, j, index + 1)
|| dfs(board, word, i, j + 1, index + 1)
|| dfs(board, word, i, j - 1, index + 1)
) {
return true;
} // backtrace.
board[i][j] = tmp;
return false;
}
}

虽然前面的方法可以AC,但考虑到最好不要更改入参的值,所以在return true前,我们还是加一个回溯代码。

 public class Solution {
public boolean exist(char[][] board, String word) {
if (board == null || board.length == 0 || board[0].length == 0) {
return false;
} int rows = board.length;
int cols = board[0].length;
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
// Check if there is a word begin from i,j.
if (dfs(board, word, i, j, 0)) {
return true;
}
}
} return false;
} public boolean dfs(char[][] board, String word, int i, int j, int index) {
int len = word.length();
if (index >= len) {
return true;
} // Check the input parameter.
if (i < 0 || i >= board.length || j < 0 || j >= board[0].length) {
return false;
} // If the current character is right.
if (word.charAt(index) != board[i][j]) {
return false;
} char tmp = board[i][j];
board[i][j] = '#'; boolean ret = dfs(board, word, i + 1, j, index + 1)
|| dfs(board, word, i - 1, j, index + 1)
|| dfs(board, word, i, j + 1, index + 1)
|| dfs(board, word, i, j - 1, index + 1); // backtrace.
board[i][j] = tmp;
return ret;
}
}

GitHub代码:

exist.java

LeetCode: Word Search 解题报告的更多相关文章

  1. 【LeetCode】79. Word Search 解题报告(Python & C++)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 回溯法 日期 题目地址:https://leetco ...

  2. LeetCode: Combination Sum 解题报告

    Combination Sum Combination Sum Total Accepted: 25850 Total Submissions: 96391 My Submissions Questi ...

  3. LeetCode & Binary Search 解题模版

    LeetCode & Binary Search 解题模版 In computer science, binary search, also known as half-interval se ...

  4. [LeetCode] Word Search II 词语搜索之二

    Given a 2D board and a list of words from the dictionary, find all words in the board. Each word mus ...

  5. [LeetCode] Word Search 词语搜索

    Given a 2D board and a word, find if the word exists in the grid. The word can be constructed from l ...

  6. 【LeetCode】Permutations 解题报告

    全排列问题.经常使用的排列生成算法有序数法.字典序法.换位法(Johnson(Johnson-Trotter).轮转法以及Shift cursor cursor* (Gao & Wang)法. ...

  7. LeetCode - Course Schedule 解题报告

    以前从来没有写过解题报告,只是看到大肥羊河delta写过不少.最近想把写博客的节奏给带起来,所以就挑一个比较容易的题目练练手. 原题链接 https://leetcode.com/problems/c ...

  8. Leetcode: word search

    July 6, 2015 Problem statement: Word Search Given a 2D board and a word, find if the word exists in ...

  9. LeetCode: Sort Colors 解题报告

    Sort ColorsGiven an array with n objects colored red, white or blue, sort them so that objects of th ...

随机推荐

  1. POJ 1163 The Triangle DP题解

    寻找路径,动态规划法题解. 本题和Leetcode的triangle题目几乎相同一样的,本题要求的是找到最大路径和. 逆向思维.从底往上查找起就能够了. 由于从上往下能够扩展到非常多路径.而从下往上个 ...

  2. exception PLS-00103: Encountered the symbol "(" when expecting one of the following:

      exception PLS-00103: Encountered the symbol "(" when expecting one of the following: Cre ...

  3. C-main函数剖析。

    对于main函数.我想不论什么一个接触到C语言的都不会陌生,可是说起main()函数有參数,你可能会产生非常多疑问了. 首先,我们来看下msdn,这个里面对main()函数有具体的说明. 在这里,我们 ...

  4. Unity3D - 资源管理

    一直没有总结过Unity的资源管理,都是随用随看文档.今天有人问起,总结一下.加深对Unity资源管理的理解. 主要參考了Unity官方文档之Resources和AssetBundle. Unity有 ...

  5. 在Ubuntu16.04 64bit上安装sublime text 3

    安装sublime text 3 根据官网上提供的安装说明 https://www.sublimetext.com/docs/3/linux_repositories.html  进行安装, 首先是 ...

  6. cocos2dx 3.2 解决输入框(TextField,TextFieldTTF) 中文乱码问题

    近期开发cocos2dx 项目,做一个小游戏.(个人喜欢用最新版本号) 没系统学习就是问题多多,遇到了非常多问题,比方全部的opengl api都必须在主线程中调用, 这让我在多线程载入方面吃了不少亏 ...

  7. ___cxa_pure_virtual&quot;, referenced from

    加入百度地图之后报这种错,解决方法:将project中的.m文件改一个成为.mm文件.

  8. 我遇到了Hibernate异常

    真是郁闷,今天想用Hibernate的实现对数据库的增删查改,但是就是报异常不断啊!呵呵,为什么?就是在主键的问题上,我用主键的生成形式是:Sequence时就报IllegalArgumentExce ...

  9. 给你的博客加上“Fork me on Github”彩带(转)

    给你的博客加上“Fork me on Github”彩带 https://www.cnblogs.com/Leo_wl/p/3608794.html https://github.blog/2008- ...

  10. 近年Recsys论文

    2015年~2017年SIGIR,SIGKDD,ICML三大会议的Recsys论文: [转载请注明出处:https://www.cnblogs.com/shenxiaolin/p/8321722.ht ...