Java 递归分词

Question

我正在尝试解决 Leetcode 上的以下问题。

"给定一个字符串 s 和一个单词字典 dict，判断 s 是否可以分割成一个或多个字典单词的空格分隔序列。

例如，给定

s = "leetcode",
dict = ["leet", "code"].

返回 true 因为“leetcode”可以被分割为“leet code”。

目前，23 次测试中有 22 次已经过去。请帮我定位问题。

public class Solution {
public boolean wordBreak(String s, Set<String> dict) {
    if(dict.contains(s) ){
        return true;
    }

    for(int i = 0; i< s.length(); i++){
        if(dict.contains(s.substring(0, i))){
            return wordBreak(s.substring(i), dict);
        }
    }
    return false;
}
}

测试失败：

"fohhemkkaecojceoaejkkoedkofhmohkcjmkggcmnami", ["kfomka","hecagbngambii","anobmnikj","c","nnkmfelneemfgcl","ah","bgomgohl","lcbjbg","ebjfoiddndih","hjknoamjbfhckb","eioldlijmmla ","nbekmcnakif","fgahmihodolmhbi","gnjfe","hk","b","jbfgm","ecojceoaejkkoed","cemodhmbcmgl","j","gdcnjj","kolaijoicbc","liibjjcini", "lmbenj","eklingemgdjncaa","m","hkh","fblb","fk","nnfkfanaga","eldjml","iejn","gbmjfdooeeko","jafogijka","ngnfggojmhclkjd","bfagnfclg ","imkeobcdidiifbm","ogeo","gicjog","cjnibenelm","ogoloc","edciifkaff","kbeeg","nebn","jdd","aeojhclmdn","dilbhl","dkk", "bgmck","ohgkefkadonafg","labem","fheoglj","gkcanacfjfhogjc","eglkcddd","lelelihakeh","hhjijfiodfi","enehbibnhfjd","gkm","ggj","ag","hhhjogk ","lllicdhihn","goakjjnk","lhbn","fhheedadamlnedh","bin","cl","ggjljjjf","fdcdaobhlhgj","nijlf","i","gaemagobjfc","dg", "g","jhlelodgeekj","hcimohlni","fdoiohikhacgb","k","doiaigclm","bdfaoncbhfkdbjd","f","jaikbciac","cjgadmfoodmba","molokllh","gfkngeebnggo","lahd ","n","ehfngoc","lejfcee","kofh moh","cgda","de","kljnicikjeh","edomdbibhif","jehdkgmmofihdi","hifcjkloebel","gcghgbemjege","kobhhefbbb","aaikgaolhllhlm","akg","kmmikgkhnn","dnamfhaf" ,"mjhj","ifadcgmgjaa","acnjehgkflgkd","bjj","maihjn","ojakklhl","ign","jhd","kndkhbebgh","amljjfeahcdlfdg","fnboolobch","gcclgcoaojc"," kfokbbkllmcd","fec","dljma","noa","cfjie","fohhemkka","bfaldajf","nbk","kmbnjoalnhki","ccieabbnlhbjmj","nmacelialookal","hdlefnbmgklo","bfbblofk" ,"doohocnadd","klmed","e","hkkcmbljlojkghm","jjiadlgf","ogadjhambjikce","bglghjndlk","gackokkbhj","oofohdogb","leiolllnjj","edekdnibja","gjhglilocif"," ccfnfjalchc","gl","ihee","cfgccdmecem","mdmcdgjelhgk","laboglchdhbk","ajmiim","cebhalkngloae","hgohednmkahdi","ddiecjnkmgbbei","ajaengmcdlbk","kgg","ndchkjdn" "heklaamafiomea","ehg","imelcifnhkae","hcgadilb","elndjcodnhcc","nkjd","gjnfkogkjeobo","eolega","lm","jddfkfbbbhia","cddmfeckheeo","bfnmaalmjdb"," fbcg","ko","mojfj","kk","bbljjnnikdhg","l","calbc","mkekn","ejlhdk","hkebdiebecf","emhelbbda","mlba","ckjmih" “哦facclfl","lgfjjbgookmnoe","begnkogf","gakojeblk","bfflcmdko","cfdclljcg","ho","fo","acmi","oemknmffgcio","mlkhk","kfhkndmdojhidg","ckfcibmnikn" "dgoecamdliaeeoa","ocealkbbec","kbmmihb","ncikad","hi","nccjbnldneijc","hgiccigeehmdl","dlfmjhmioa","kmff","gfhkd","okiamg","ekdbamm"," fc","neg","cfmo","ccgahikbbl","khhoc","elbg","cbghbacjbfm","jkagbmfgemjfg","ijceidhhajmja","imibemhdg","ja","idkfd","ndogdkjjkf" "fhic","ooajkki","fdnjhh","ba","jdlnidngkfffbmi","jddjfnnjoidcnm","kghljjikbacd","idllbbn","d","mgkajbnjedeiee","fbllleanknmoomb","lom"," kofjmmjm","mcdlbglonin","gcnboanh","fggii","fdkbmic","bbiln","cdjcjhonjgiagkb","kooenbeoongcle","cecnlfbaanckdkj","fejlmog","fanekdneoaammb","maojbcegdamn","bcmanmjdeabdo" "amloj","adgoej","jh","fhf","cogdljlgek","o","joeiajlioggj","oncal","lbgg","elainnbffk","hbdi","femcanllndoh"," ke","hmib","nagfahhljh","ibifdlfeechcbal","knec","oegfcghlgalcnno","abiefmjldmln","mlfglgni","jkofhjeb","ifjbneblfldjel","nahhcimkjhjgb","cdgkbn","nnklfbeecgedie" ” “gmllmjbo dhgllc","hogollongjo","fmoinacebll","fkngbganmh","jgdblmhlmfij","fkkdjknahamcfb","aieakdokibj","hddlcdiailhd","iajhmg","jenocgo","embdib","dghbmljjogka","bahcggjgmlf" ,"fb","jldkcfom","mfi","kdkke","odhbl","jin","kcjmkggcmnami","kofig","bid","ohnohi","fcbojdgoaoa","dj"," ifkbmbod","dhdedohlghk","nmkeakohicfdjf","ahbifnnoaldgbj","egldeibiinoac","iehfhjjjmil","bmeimi","ombngooicknel","lfdkngobmik","ifjcjkfnmgjcnmi","fmf","aoeaa","an" "ffgddcjblehhggo","hijfdcchdilcl","hacbaamkhblnkk","najefebghcbkjfl","hcnnlogjfmmjcma","njgcogemlnohl","ihejh","ej","ofn","ggcklj","omah","hg"," obk","giig","cklna","lihaiollfnem","ionlnlhjckf","cfdlijnmgjoebl","dloehimen","acggkacahfhkdne","iecd","gn","odgbnalk","ahfhcd","dghlag" "bchfe","dldblmnbifnmlo","cffhbijal","dbddifnojfibha","mhh","cjjol","fed","bhcnf","ciiibbedklnnk","ikniooicmm","ejf","ammeennkcdgbjco"," jmhmd","cek","bjbhcmda","kfjmhbf","chjmmnea","ifccifn","naedmco","iohchafbega","kjejfhbco","anlhhhhg"]

output: false. Expected: true.

Answer 1

你应该测试到最后。您的方法在第一部分匹配后立即返回 (return wordBreak ... )，但可能与其他起始词有不同的匹配。 例如。您测试 aabaa 并且您的 dict 是 [aa, aab] 那么您的方法将返回 false 因为首先找到 aa 并且 咩 不在dict 内。

您必须测试所有组合。将您的方法更改为

public boolean wordBreak(String s, Set<String> dict) {
    if (dict.contains(s)) {
        return true;
    }

    for (int i = 0; i < s.length(); i++) {
        if (dict.contains(s.substring(0, i))) {
            //Here is the modification:
            if (wordBreak(s.substring(i), dict))
                return true;
        }
    }
    return false;
}

你看，它只在 wordBreak 的递归调用成功时返回，而不是在所有情况下。