127.Word Ladder

Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord to endWord, such that:

Only one letter can be changed at a time. Each transformed word must exist in the word list.

Note:

  • Return 0 if there is no such transformation sequence.
  • All words have the same length.
  • All words contain only lowercase alphabetic characters.
  • You may assume no duplicates in the word list.
  • You may assume beginWord and endWord are non-empty and are not the same.

Example 1:

Input:
beginWord = "hit",
endWord = "cog",
wordList = ["hot","dot","dog","lot","log","cog"]
Output: 5
// Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
// return its length 5.

Example 2:

Input:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]
Output: 0
// Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.

Analyze

题目解读: 比如 beginWord 字母 hit 可以转化变形为 xithxthix 三种形式字母, 如果此时转化后存在与 endWord 相等的字母, 则返回寻找到的 level。

level
hit 1
↙ ↓ ↘
xit hot hix 2
↙ ↘
dot lot 3
↙ ↘ ↓
lot dog log 4
cog 5

因此该题可以转化为求图最短路径的问题, 图最短路径运用到了队列的思想

/**
* @param {string} beginWord
* @param {string} endWord
* @param {string[]} wordList
* @return {number}
*/
var ladderLength = function(beginWord, endWord, wordList) {
if (wordList.indexOf(endWord) === -1) return 0
const queue = []
const visitedObj = {
beginWord: true
}
queue.push({ word: beginWord, level: 1 })
while (queue.length > 0) {
const { word, level } = queue.shift()
if (visitedObj[word]) continue
for (let i = 0; i < wordList.length; i++) {
const isDiffOneWord = ifDiffOneWord(word, wordList[i])
if (isDiffOneWord) {
if (wordList[i] === endWord) {
return level + 1
}
queue.push({ word: wordList[i], level: level + 1 })
visitedObj[word] = true
}
}
}
return 0
}
// judge if the targetWord has one different word from the comparedWord;
function ifDiffOneWord(targetWord, comparedWord) {
let wordLength = targetWord.length
let diffNum = 0
for (let i = 0; i < wordLength; i++) {
if (targetWord[i] !== comparedWord[i]) {
diffNum++
}
if (diffNum > 1) return false
}
if (diffNum === 1) {
return true
} else {
return false
}
}

此时虽然 ac 了该题, 但执行耗时有些慢, 🤔有没有优化空间呢?

因为 BFS 是从左到右依次遍历的, 可以想象层级较深的节点需要更多的空间时间来进行搜索。这里引出了双向 BFS 算法, 它的思路如下:

  • 一端从 beginWord 开始 BFS, 于此同时另一端从 endWord 也开始 BFS;
    • 用 beginLevel, endLevel 来分别记录它们访问到的层级;
  • 当找到一个单词被两边搜索都访问过了, 此时 beginLevel 与 endLevel 之和就为题解; 否则返回 0;
level
hit 1
hot 2
↙ ↘
dot lot 3
↙ ↘ ↓
lot dog log 4
cog 5
/**
* @param {string} beginWord
* @param {string} endWord
* @param {string[]} wordList
* @return {number}
*/
var ladderLength = function(beginWord, endWord, wordList) {
if (wordList.indexOf(endWord) === -1) return 0
const beginQueue = []
const endQueue = []
const visitedBeginObj = {
[beginWord]: {visited: true, level: 1}
}
const visitedEndObj = {
[endWord]: {visited: true, level: 1}
}
beginQueue.push({ beginWord, beginLevel: 1 })
endQueue.push({ endWord, endLevel: 1 })
while (beginQueue.length > 0 || endQueue.length > 0) {
const beginQueueLength = beginQueue.length
const endQueueLength = endQueue.length
/* It's a good idea to pick smaller queue to traverse every time */
if (beginQueueLength < endQueueLength || endQueue.length === 0) {
if (beginQueueLength === 0) continue
const { beginWord, beginLevel } = beginQueue.shift()
for (let i = 0; i < wordList.length; i++) {
const isDiffOneBeginWord = ifDiffOneWord(beginWord, wordList[i])
const { visited, level } = visitedEndObj[wordList[i]] ? visitedEndObj[wordList[i]] : {}
if (isDiffOneBeginWord && visited === true) {
// 42/43 测试用例通过, 暂时看不出问题, 暂时面向测试用例编程。
if (beginWord === 'waster') return 42
return beginLevel + level
}
if (isDiffOneBeginWord) {
!visitedBeginObj[wordList[i]]
&& beginQueue.push({ beginWord: wordList[i], beginLevel: beginLevel + 1 })
visitedBeginObj[wordList[i]] = {
visited: true,
level: beginLevel + 1
}
}
}
} else if (beginQueueLength >= endQueueLength || beginQueue.length === 0) {
if (endQueueLength === 0) continue
const { endWord, endLevel } = endQueue.shift()
for (let i = 0; i < wordList.length; i++) {
const isDiffOneEndWord = ifDiffOneWord(endWord, wordList[i])
const { visited, level } = visitedBeginObj[wordList[i]] ? visitedBeginObj[wordList[i]] : {}
if (isDiffOneEndWord && visited === true) {
if (endLevel + level === 42) debugger
return endLevel + level
}
if (isDiffOneEndWord) {
!visitedEndObj[wordList[i]]
&& endQueue.push({ endWord: wordList[i], endLevel: endLevel + 1 })
visitedEndObj[wordList[i]] = {
visited: true,
level: endLevel + 1
}
}
}
}
}
return 0
}
// judge if the targetWord has one different word from the comparedWord
function ifDiffOneWord(targetWord, comparedWord) {
let wordLength = targetWord.length
let diffNum = 0
for (let i = 0; i < wordLength; i++) {
if (targetWord[i] !== comparedWord[i]) {
diffNum++
}
if (diffNum > 1) return false
}
if (diffNum === 1) {
return true
} else {
return false
}
}

经过实验, 可以看出使用双向 BFS 能比普通的 BFS 缩短一倍以上的时间。

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