介绍

题目链接

方法一

这是我的第一种实现方法，时间复杂度很高，没有通过全部的测试用例，具体看代码就可以了

代码

Solution.java

import java.util.ArrayList;
import java.util.List;class Solution {public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {//这是最后的返回结果List<List<String>> ans = new ArrayList<List<String>>();//表示最短的序列长度， 初始值设置为最大Integer[] le = {Integer.MAX_VALUE};//la表示其中一个结果集List<String> la = new ArrayList<String>();//加入第一个元素la.add(beginWord);//定义一个数组变量来随时表示字典中的字是否被使用过//多给一个元素，保证下面的程序理解起来简单Boolean[] btest = new Boolean[wordList.size() + 1];//定义一个循环，遍历wordList，从beginWord开始查找，在列表中查找和beginWork相差一个单词的元素，那就把这个元素加到一个la中//并将该元素相应的位置设置为false,表示该元素不能再被读取，然后继续循环，//直到元素与endWord相同，或者是查找完所有的元素，或者是没有元素可以与之匹配，//如果当前位置已经大于le，则直接进行回溯，进行下一次查找//如果是与endWord相同，则进行下一次查找，但是不能和上一个的结果集相同   这里要记录le//如果是查找完所有的元素，则返回空列表就可以了//如果是没有元素与之匹配，这里应该是进行回溯，在这里可以考虑使用递归进行回溯，否则直接回溯会很麻烦//优化问题，这里要找的是最短的序列，因此要记录一个最短的数量//for(int i=0; i<btest.length; i++) {btest[i] = true;}Integer let = 0;//goBack(beginWord, endWord, wordList, ans, le, let, la, btest);for(int i=1;i<wordList.size()+2;i++) {le[0] = i;goBack(beginWord, endWord, wordList, ans, le, let, la, btest);if(ans.size() != 0) {break;}}//排除掉所有不合适的元素for(int i=0; i<ans.size();i++){//因为有初始元素，所以要+2while(i<ans.size() && ans.get(i).size() > (le[0]+1)) {ans.remove(i);}}return ans;}/*** 定义一个回溯函数（递归函数）* 该递归函数应该查找出需要的结果集* * 对函数的结构进行说明，其中应该有一个循环，来进行循环的查找* @param beginWord 开始字符* @param endWord    结束字符* @param wordList  字典，所有的中间结果* @param ans      最终结果* @param le        表示最短的序列长度， le应该是引用，可以进行修改* @param let        表示当前是第几个元素，使用引用* @param la      表示其中一个结果集* @param btest        表示该元素是否被使用过* */void goBack(String beginWord, String endWord, List<String> wordList, List<List<String>> ans, Integer[] le, Integer let, List<String> la, Boolean[] btest) {//1.首先let自增，判断是否超出了最短序列，超出则直接返回let++;if(let>le[0]) {return;}//2.循环开始---for(int i=0;i<wordList.size(); i++) {//3.#---从beginWord开始查找，在列表中查找和beginWork相差一个单词的元素,在该元素为true的情况下//如果找到了这样一个元素{if(btest[i] && subOne(beginWord,wordList.get(i)) ) {//4.那就把这个元素加到一个la中la.add(wordList.get(i));//5.---btest将该元素相应的位置设置为false,表示该元素不能再被读取btest[i] = false;//6.---如果是与endWord相同，则进行回溯，要记录le，//这里(如果le更小，或者相等都可以)//  然后要将la放到ans中，//表示找到了其中一个结果这里已经找到结果，这里需要返回return// 返回(return)之前需要做的操作:首先la中加入的元素删除， 然后btest改为true,最后let--// 也就是认为还没有找到结果，继续进行查找if(wordList.get(i).equals(endWord)) {if(let < le[0]) {le[0] = let;}//重新定义一个laaList<String> laa = new ArrayList(la);ans.add(laa);//System.out.println(laa);la.remove(la.size()-1);//删除最后一个元素btest[i] = true;//let[0]--;//进行查找下一个return;}//7.---这里就是与不endWord相同，表示需要查找下一个元素，将beginWord设置为当前的结果，调用自身递归查询，进行递归即可String beginWordIn = wordList.get(i);goBack(beginWordIn, endWord, wordList, ans, le, let, la, btest);//8.---递归返回了，说明该路径已经查找完毕，必须弹出当前点，找另外一个点替换当前点，这里做弹出当前点的操作就可以//.---弹出当前点需要做的操作:首先la中加入的元素删除， 然后btest改为truela.remove(la.size()-1);//删除最后一个元素btest[i] = true;//let[0]--;    //这里不用减少，因为还是在同函数中}//#   }}//9.循环跳出---如果是没有元素与之匹配，直接进行回溯就可以，// 因为之前弹出了当前点，所以这里不用重复弹出//  函数返回之前要减少//let[0]--;return;}/*** 判断当前这两个字符串是否相差一个元素，相等的情况不做考虑* @param one 第一个字符串* @param two 第二个字符串* @return*/boolean subOne(String one, String two) {//表示当前有几个元素不同int sum = 0;for(int i=0;i<one.length();i++){//如果元素不同，则计数器加一if(one.charAt(i) != two.charAt(i)) {sum++;if(sum > 1) {break;}}}//如果只有一个不同，返回trueif(sum ==1 ) {return true;}else {return false;}}
}

Test.java

import java.util.ArrayList;
import java.util.List;public class Test {public static void main(String[] args) {Solution s = new Solution();String beginWord = "hit";String endWord = "cog";List<String> wordList = new ArrayList<String>();String[] a = {"hot","dot","dog","lot","log","cog"};for(int i=0;i<a.length;i++) {wordList.add(a[i]);}System.out.println(s.findLadders(beginWord, endWord, wordList));System.out.println(s.subOne("dot", "log"));;Integer i = new Integer(666666);fun(i);System.out.println(i);}public static void fun(Integer i) {i = 1;}
}

方法二

算法介绍

官方的解答
算法是参考官方的解答，使用的方法是图的广度优先遍历，在遍历的过程中记录路径，因为要找的是所有的最短路径，所以要做一个记录，然后遍历这些路径，看路径的最后一个点是否是终点，当然在前面应该判断，如果字典中没有终点，直接返回空就可以了。

在这个过程中需要注意的问题

注意一下问题

Integer i1 = 200;
Integer i2 = 200;#结果是false
System.out.println(i1 == i2);#结果是true
System.out.println(i1.equals(i2));

为什么是这样的，请详细查看java关于Integer的说明，查看Integer的源代码进行理解，在此只给出一个范围。 -128~+127 之间都是true，超出这个范围的是false和true。本人在这里栽了，这种问题是比较难以发现的问题，想多了解请查看文档。

代码

Solution.java

import java.util.ArrayList;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Queue;
/*** 此种解决方案是使用广度优先算法* @author 123**/
class Solution {public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {//这是最后的返回结果List<List<String>> ans = new ArrayList<List<String>>();//建立序号和图的序号//由字符串查找点Map<String, Integer> sitm = new HashMap<String, Integer>();//由序号查找字符串List<String> sitl = new ArrayList<String>();for(int i=0;i<wordList.size(); i++) {String temp = wordList.get(i);//这里判断一下更好，避免有重复的数据if(!sitm.containsKey(temp)) {sitm.put(temp, i);sitl.add(temp);}}//如果终点不在字典中，直接退出if(!sitm.containsKey(endWord)) {return ans;}//将起点加入sitm.put(beginWord, wordList.size());sitl.add(beginWord);//将图定义和实现出来//List<List<Integer>>List<List<Integer>> mapl = new ArrayList<List<Integer>>();//图for(int i=0;i<sitl.size(); i++) {List<Integer> temp = new ArrayList<Integer>();mapl.add(temp);}//for(int i=0;i<sitl.size(); i++) {for(int j=i+1;j<sitl.size(); j++) {if(subOne(sitl.get(i), sitl.get(j))){//建立一条无向边mapl.get(i).add(j);mapl.get(j).add(i);}}}//首先应该定义一个队列//QueueQueue<List<Integer>> q = new LinkedList<List<Integer>>();//定义权重List<Integer> weight = new ArrayList<Integer>();for(int i=0;i<sitl.size();i++) {weight.add(Integer.MAX_VALUE);}//将开始节点的权重设置为0weight.set(weight.size()-1, 0);//定义result，记录过程路径List<Integer> result = new ArrayList<Integer>();result.add(sitm.get(beginWord));q.add(result);while(!q.isEmpty()) {List<Integer> one = q.poll();//如果最后一个元素是endWordInteger last = one.get(one.size()-1); //这是路径中的最后一个元素if(one.size() == 11) {int sfsf;sfsf = 1;}if(last.equals(sitm.get(endWord))) {//那么将该路径放入到结果集List<String> temp = new ArrayList<String>();for(int i=0;i<one.size(); i++){temp.add(sitl.get(one.get(i)));}ans.add(temp);}else {//否则就查找下一个元素List<Integer> visit = new ArrayList<Integer>(mapl.get(last));for(int i=0;i<visit.size();i++) {Integer temp = visit.get(i);if(weight.get(temp) >= weight.get(last)+1) {//更改权重weight.set(temp, weight.get(last)+1);List<Integer> temp2 = new ArrayList<Integer>(one);temp2.add(temp);//debug//debugq.add(temp2);}}}}//进行遍历return ans;}/*** 判断当前这两个字符串是否相差一个元素，相等的情况不做考虑* @param one 第一个字符串* @param two 第二个字符串* @return*/boolean subOne(String one, String two) {//表示当前有几个元素不同int sum = 0;for(int i=0;i<one.length();i++){//如果元素不同，则计数器加一if(one.charAt(i) != two.charAt(i)) {sum++;if(sum > 1) {break;}}}//如果只有一个不同，返回trueif(sum ==1 ) {return true;}else {return false;}}
}

Test.java

import java.util.ArrayList;
import java.util.List;public class Test {public static void main(String[] args) {Solution s = new Solution();String beginWord = "cet";String endWord = "ism";List<String> wordList = new ArrayList<String>();String[] a = {"kid","tag","pup","ail","tun","woo","erg","luz","brr","gay","sip","kay","per","val","mes","ohs","now","boa","cet","pal","bar","die","war","hay","eco","pub","lob","rue","fry","lit","rex","jan","cot","bid","ali","pay","col","gum","ger","row","won","dan","rum","fad","tut","sag","yip","sui","ark","has","zip","fez","own","ump","dis","ads","max","jaw","out","btu","ana","gap","cry","led","abe","box","ore","pig","fie","toy","fat","cal","lie","noh","sew","ono","tam","flu","mgm","ply","awe","pry","tit","tie","yet","too","tax","jim","san","pan","map","ski","ova","wed","non","wac","nut","why","bye","lye","oct","old","fin","feb","chi","sap","owl","log","tod","dot","bow","fob","for","joe","ivy","fan","age","fax","hip","jib","mel","hus","sob","ifs","tab","ara","dab","jag","jar","arm","lot","tom","sax","tex","yum","pei","wen","wry","ire","irk","far","mew","wit","doe","gas","rte","ian","pot","ask","wag","hag","amy","nag","ron","soy","gin","don","tug","fay","vic","boo","nam","ave","buy","sop","but","orb","fen","paw","his","sub","bob","yea","oft","inn","rod","yam","pew","web","hod","hun","gyp","wei","wis","rob","gad","pie","mon","dog","bib","rub","ere","dig","era","cat","fox","bee","mod","day","apr","vie","nev","jam","pam","new","aye","ani","and","ibm","yap","can","pyx","tar","kin","fog","hum","pip","cup","dye","lyx","jog","nun","par","wan","fey","bus","oak","bad","ats","set","qom","vat","eat","pus","rev","axe","ion","six","ila","lao","mom","mas","pro","few","opt","poe","art","ash","oar","cap","lop","may","shy","rid","bat","sum","rim","fee","bmw","sky","maj","hue","thy","ava","rap","den","fla","auk","cox","ibo","hey","saw","vim","sec","ltd","you","its","tat","dew","eva","tog","ram","let","see","zit","maw","nix","ate","gig","rep","owe","ind","hog","eve","sam","zoo","any","dow","cod","bed","vet","ham","sis","hex","via","fir","nod","mao","aug","mum","hoe","bah","hal","keg","hew","zed","tow","gog","ass","dem","who","bet","gos","son","ear","spy","kit","boy","due","sen","oaf","mix","hep","fur","ada","bin","nil","mia","ewe","hit","fix","sad","rib","eye","hop","haw","wax","mid","tad","ken","wad","rye","pap","bog","gut","ito","woe","our","ado","sin","mad","ray","hon","roy","dip","hen","iva","lug","asp","hui","yak","bay","poi","yep","bun","try","lad","elm","nat","wyo","gym","dug","toe","dee","wig","sly","rip","geo","cog","pas","zen","odd","nan","lay","pod","fit","hem","joy","bum","rio","yon","dec","leg","put","sue","dim","pet","yaw","nub","bit","bur","sid","sun","oil","red","doc","moe","caw","eel","dix","cub","end","gem","off","yew","hug","pop","tub","sgt","lid","pun","ton","sol","din","yup","jab","pea","bug","gag","mil","jig","hub","low","did","tin","get","gte","sox","lei","mig","fig","lon","use","ban","flo","nov","jut","bag","mir","sty","lap","two","ins","con","ant","net","tux","ode","stu","mug","cad","nap","gun","fop","tot","sow","sal","sic","ted","wot","del","imp","cob","way","ann","tan","mci","job","wet","ism","err","him","all","pad","hah","hie","aim","ike","jed","ego","mac","baa","min","com","ill","was","cab","ago","ina","big","ilk","gal","tap","duh","ola","ran","lab","top","gob","hot","ora","tia","kip","han","met","hut","she","sac","fed","goo","tee","ell","not","act","gil","rut","ala","ape","rig","cid","god","duo","lin","aid","gel","awl","lag","elf","liz","ref","aha","fib","oho","tho","her","nor","ace","adz","fun","ned","coo","win","tao","coy","van","man","pit","guy","foe","hid","mai","sup","jay","hob","mow","jot","are","pol","arc","lax","aft","alb","len","air","pug","pox","vow","got","meg","zoe","amp","ale","bud","gee","pin","dun","pat","ten","mob"};for(int i=0;i<a.length;i++) {wordList.add(a[i]);}System.out.println(s.findLadders(beginWord, endWord, wordList));System.out.println("*----------------------------------------------");System.out.println("**************************************************");//其他测试System.out.println("==============================");System.out.println(s.subOne("dot", "log"));;Integer i = new Integer(666666);fun(i);System.out.println(i);}public static void fun(Integer i) {i = 1;}
}

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leetcode 126. 单词接龙 II

leetcode 126题

介绍

方法一

代码

Solution.java

Test.java

方法二

算法介绍

在这个过程中需要注意的问题

代码

Solution.java

Test.java

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