标签:bfs cas eth javascrip val hit script turn 字符
126. Word Ladder II
Given two words (beginWord and endWord), and a dictionary‘s word list, find all shortest transformation sequence(s) from beginWord to endWord, such that:
Note:
Example 1:
Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"] Output: [ ["hit","hot","dot","dog","cog"], ["hit","hot","lot","log","cog"] ]
Example 2:
Input: beginWord = "hit" endWord = "cog" wordList = ["hot","dot","dog","lot","log"] Output: [] Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.
题意:给出开始单词和结束单词,从单词列表中找出单词使得从开始单词到结束单词形成接龙,每次只允许替换一个字符,求出所有情况
代码如下:
‘use strict‘; var findLadders = function(start, end, dict) { // 常量 var A_CODE = ‘a‘.charCodeAt(0); var WORD_COUNTS = 26; var wordLength = start.length; var results; var currents, next; var isFounded = false; var dictSet = new Set(dict); // 判断一次字符转换是否有效 var isValid = function(from, to) { var i = 0, c = 0; while (i < wordLength) { if (from.charCodeAt(i) !== to.charCodeAt(i)) { ++c; } ++i; } return (c === 1); }; // 字符替换 var replacedWord = function(word, idx, chCode) { var newStr = word.substr(0, idx) + String.fromCharCode(chCode) + word.substr(idx + 1); return newStr; }; // If its only one step from start to end. if (start === end || isValid(start, end)) { return [[start, end]]; } else if (!dictSet.has(end)) { return []; } results = []; var startSet = new Set([start]); var endSet = new Set([end]); var startPath = [[start]]; var endPath = [[end]] var isReversing = false; var isConnected = false; // Use to decide whether use all word possible or use all dice word. var wordCombinations = WORD_COUNTS * wordLength; var dictComputations; // Determine current paths. var currentPaths; var currentLength; var currentSet; var pathLength; var nextPaths; var currentPath, currentWord, targets, target, tmpPath; var i, j, k; // 初始化 currentPaths = startPath; currentSet = startSet; currentLength = currentPaths.length; while (currentLength > 0) { nextPaths = []; // 从currentSet中删除key targets = currentSet.keys(); for (target of targets) { dictSet.delete(target); } currentSet.clear(); dictComputations = dictSet.size * wordLength; // Decide whether to use dict iteration of word replaces. if (dictComputations < wordCombinations) { // If iteration though dict needs less compares, iterate it. for (i = 0; i < currentLength; ++i) { currentPath = currentPaths[i]; currentWord = currentPath[currentPath.length - 1]; targets = dictSet.keys(); for (target of targets) { if (isValid(currentWord, target)) { tmpPath = currentPath.slice(); tmpPath.push(target); nextPaths.push(tmpPath); currentSet.add(target); } } } } else { for (i = 0; i < currentLength; ++i) { currentPath = currentPaths[i]; currentWord = currentPath[currentPath.length - 1]; for (j = 0; j < wordLength; ++j) { for (k = 0; k < WORD_COUNTS; ++k) { target = replacedWord(currentWord, j, A_CODE + k); if (dictSet.has(target)) { tmpPath = currentPath.slice(); tmpPath.push(target); nextPaths.push(tmpPath); currentSet.add(target); } } } } } if (isReversing) { endPath = nextPaths; } else { startPath = nextPaths; } if (startSet.size > endSet.size) { targets = endSet.keys(); currentSet = startSet; } else { targets = startSet.keys(); currentSet = endSet; } for (target of targets) { if (currentSet.has(target)) { isConnected = true; break; } } if (isConnected) { break; } else { // 取小 isReversing = startPath.length > endPath.length ? true : false; currentSet = isReversing ? endSet : startSet; currentPaths = isReversing ? endPath : startPath; currentLength = currentPaths.length; } } if (isConnected) { currentLength = startPath.length; pathLength = endPath.length; // Reverse endPaths. for (j = 0; j < pathLength; ++j) { endPath[j].reverse(); } for (i = 0; i < currentLength; ++i) { currentPath = startPath[i]; currentWord = currentPath[currentPath.length - 1]; if (!endSet.has(currentWord)) { continue; } for (j = 0; j < pathLength; ++j) { target = endPath[j]; if (currentWord === target[0]) { tmpPath = currentPath.concat(target.slice(1)); results.push(tmpPath); } } } } return results; };
标签:bfs cas eth javascrip val hit script turn 字符
原文地址:https://www.cnblogs.com/xingguozhiming/p/10909234.html