给你一份『词汇表』（字符串数组） words 和一张『字母表』（字符串） chars。
假如你可以用 chars 中的『字母』（字符）拼写出 words 中的某个『单词』（字符串），那么我们就认为你掌握了这个单词。
注意：每次拼写时，chars 中的每个字母都只能用一次。
返回词汇表 words 中你掌握的所有单词的 长度之和。
示例 1：
输入：words = ["cat","bt","hat","tree"], chars = "atach"
输出：6
解释：
可以形成字符串 "cat" 和 "hat"，所以答案是 3 + 3 = 6。
示例 2：
输入：words = ["hello","world","leetcode"], chars = "welldonehoneyr"
输出：10
解释：
可以形成字符串 "hello" 和 "world"，所以答案是 5 + 5 = 10。
提示：
1 <= words.length <= 1000
1 <= words[i].length, chars.length <= 100
所有字符串中都仅包含小写英文字母
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/find-words-that-can-be-formed-by-characters
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
package leetCodeTest;
import java.util.*;
public class 拼写单词 {
public static void main(String[] args) {
String words[] = new String[]{"jdimdfzbygjgdrsisgblztowvwzewvrwuizaihnctoonlykrbzncqvbzjahxvyswdzcbwaqliiyzbddtsxrwb","ybdhccgznflezzvifoshbatwedohwwhyajylprerarbsjcll","febacfxkrwnkyoyhykzbfuueohtkvqjkrmclmiinmpwhumaummpxjqmvqnvlwkmiizewfnqnmjqmnxacrhgxodyglvzyyii","tcymwrytgfnbggzqptaqwzpkjhleisuwsfzhna","rdvuvrrqerihadnurdfkzfiwtodnzcwvrmuwkyjhgkdjtnuwlclnhttidukyhfdzozix","mjoryzsbvkqlmtmcmsjljloabfaliecpmmiumq","iwzpczbknmqrankiypxjxthlzsvhchpqv","dqpgxyspa","igsvidvyonqyprvkoaphuvqaxcceefckypojgfdiyubagihkztnhindjcdpolt","iamvwzvdbfonuurfdzcnvohqrpyxgmjvbyiyovvccsilkblgpojmwvhdsewldgzvdlnecfudjjkmgmil","lbtjpakidwtavvtllcpekgcmqigmrfkvfjyzbxlpxtnklqrfoasnocjnlhzafcjqdjqgmuardsifyrlkvrndat","tfqdhgvduounzgbdnipofenkqzqrlntcppwkk","dwtuwpqugxvkjnhxvpxsdxyeurkldhxnpmicxmntiokhcclptbbbamzcrwachvzi","pdnmdakvxnwdmibwqclesycgomztyowtjskvwiwuaumwubrmkli","sxabmaciwtbbtwwawfhqisthjqxsxslsptcxqgsrzxiirxmzzmoulqmwzwatkrjqgpelkenulstuxwzjno","enhtapuuyildahjyryubmjqdhbtfhjzhwppqbzriytu","momfgjhyxzpsp","hhdyxizwufxjewqvbforsauqgexpqgrhbmycwcaxbwldljuhclpyfawbhypinsfjepgded","cgowwwhvlaziuptfsxqygxxnqxbregcwaodyklkkywpdtofrupmeeaeoemmpdwltdzbvebuzfumfdbdqu","xspywgpzetvfdzeyterumsweldzsdarwshssaxlfgqqlckcyeealjqdfkjdncoivvgkmcvaod","ubuubimfvxvplgikclwilovjdkns","luzrzwfwzlunkwlvpjqpwckboxhgufsjlpmklxxtiuvlvaydkdvqumhiospabiemygntpfotocufttzzewkbxf","mqqktykhjxtebccvpwttybgvkkxndvxfagdnjmbezlmdgw","opcobbidwttpeisekgmowtgcozeinvfliqbkbrtkvwxdobavjyrknsadjijwybtbayfytxeotfp","c","cefqqjft","ijavryhszfeevxvrcubkvpobxajzfjhnyffkdnucigmisvalwfngwahy","pzfkcxeehzytxrtbdqqcoaccnqiynsjypfrcablnznvmthga","tlekyalvnyxsvr","jsahgkicszfcheyqafvflufycicegxclqpbmvmakjbhtuwhcxerxwspgzsrubfqblbhavlwfhx","tvzrvrsnjetkoxbliawcvzzakpmhhvnjubczcpgobuqjqffsrdxtxjarzgsuryajuatcyfbroginvjhlauzbrkhssf","dqbrdlmkhkrdzecilmfuavrphuxbqjo","pmdpypebaobezmcfazewjbabjqjvrefyravwrvneyfpesbwnbbymuvovjvqgfwmzcsweifbpkuzicpjsj","wfobqlvkmehlrh","soormzqclvrrhftinorqqtfprxqufqsadalczqgbdnirkohqvzucsuunzjvbcfrryzztgy","lfpvzygoenywdaddtsnrtmdhqnrkuyydwbwmzbwxzmoxwxvyttomlqdsekhyoaltxjhdtggwxrueelgh","sikbegcpidcsnbierikpueovdqaxzis","bchfnjcxqkfavrpotyghvhdzfsnigpgeuen","igcpbhexbrhqgvzbjuvgehqrntbdfluouysupmtykobvkxzhumwxtntmbnyditvjmgoqtbiabkzfx","xlgxysycbrskihoxiolefjhgdluugsuvvlhxlbqxzcqgqoxrmhmrylpfbarrplqgpvep","oivjponqitutsacujnlmwotbeinwfygnzdzqet","yktuffealtkznzvccxuzejpruspfldidkheondsmdbgwrabkmwgtbguayqhyiynfjfqgbpxoizlhottxeqbauj","sxghynxeletecuqbafmqmlmiwlaenuujfipfkeylcyfjxkznpkc","gxekcfwgn","zitnbxgfpnzdoaahksuajugypfdlzwzadmksgtuqoevnqjkwoybabozgovuzlazwlopcuofwnqpkcpyupnh","oydlvwonusmqypnlpwhdakzdoft","kphotdvwvjypd","cdcwreimmvstaytgnlwrdzumkilseqrorlsgzmmogjixszokypjalqmexhjesknsnpgjowypsno","bbfbbepyutonbnmtbmmdrgxnhoifxluendcayelykgoglnrucegkhoczbajaobdggmusdbdllz","tmelbseagubuvjjvpzkyvbxcyxzoxjvhcqwmesdwi","twbedagexfuyyghuqxgwxmtbckchmiknghoaswufihzcmcdkyiqphjwpgpqsavoiiejfoqoucmvcahxmshu","aqnipfzpclcsyuecagdfho","cfcigrykqaeclkckntqkioexxvzvqfyirgqiwsbamtigxewdgpbhxsrwiofxzakvofqzbznbkfpfqpnns","uoyzgkluftocnmbdrpyrmaywyyevbfcebzfxqljgigcgfmoeznqpkbmiwmziqlmdealiknlfzdwbadyjunth","pniwsxvfxlsfhkhijuwcdvzahgzjmxdjpggcmynlfeitrggpgfudhfsswqaxapmkalrmiikrdcqyovqqyiasc","hjoff","zchyafrvstjxrpcrjtxrzqtgemtnbioizjvrsbmmwsqfcqtovk","rqopixeyjdtwekhdilymtmbvpllrbcfkfojwyzgvhlhxvflywzqiiqlnwooudmwjtfikdgecxsbwxnmmyaxkqyxbqsgn","jrjjoqkpqfjslcklltztyhkmrqmlmvjczidsswyhmyowxuunogxomhxtireakdfheffjaarcmonvxxuhvlnpujenrvjpsdrngsif","oluqeooumhfdizvbmzzxznccypdfqeibvhhbbknqcvdryuiqzmheivglczcxcneypm","odiyjnibfvnmojenjmtnawnnsatslheoqliyqtsmrxojxlsqep","xodhotsxujcannmgvmszwnpvlyxrvfntk","pal","pfuaarkhepsjjtljpdhne","cmravobrplecnohwhfcxbgsmkqkqjmdrpz","rhwlfbsjnoileeuatpwpzxsqssfxbakasvvhrnfvsnji","cprpqsoavnarqcyjhuxeeapjwkbrignbhgkufpdbentazdgfrjksrmbvcbwypcrh","ouxsbfsvuihlhjcsiccibxvxtpwuzhrthhotbinvclelxhujtxqnpzoylzatzgdvawmfjwychqeopcrdjnejdtfkdign","zhoel","jwcdwcwlljltoiepktkflngdsqeagqwowkbopjzwbruhzntusrixinwvf","amn","zpcniceupkondvpyjbckasmevmoxquumalzrnhpxskxugik","cvefdfpkcctzpipaodrygoaoesgvihhhthygs","bkohexejlvylwkrykmmcgcf","olpytsjqlkdpwi","qsidlbvlccnjtyrupfxprlwhcynoyxlbrxjojkiqrsawbaumnzkdkzchommc","ixnpbwnjtpqczoybczviqfbjtuxjdibzgispvxdnjppranzldirjcxut","zlzvegbitlzahbmkectimwpbaletlxpcotygzvwridoiswxiuawtz","fhptimdesuoqzjiblcwnhhbaxptzraksykvfpunxfufrwcmvqnwjocoiy","aiyorgqnkvdopapcjpenkymhnynpkzyntylozbydnpibdrpbvcutpxudrcsizoyk","nmuyeikwomlcripugwjypkfvbvizvrmndhsacmdocjjdhtnqvuxmgrcfeczuxsa","geulhswpnmyweuswggltxsguxnnvecovqxvteiuadxgdksmjtitwcpzmbxicwvclkndttsduxb","tbdazbskpejjkquorhfdafdvlgglefjhqguriqaefpvhcqscfhtvkimqvjzxtaf","ldnnyerfedgsffdhpbmbwuubrzsjibafxqccryuljvusadrlfxqw","meylutslakssulbpvfonrfmvugblinwzrfgopxafnwwdcuufjuaimxrbheehjndxbiohskmvjrmpcewvvvrbaqbkfffgbb","ygjmeswiarqiwkpuzjsfnbabtvgapaoazzkuafcpuuavzmvvhcnwedpgbizznkygooedxjggbdkrgkcyaautpvwwnvlfwikyl","mjgnewabmklmzwazyhcdoyqowpgrdcnuzwoha","nljceveqjwgwmylmbgkawzjdvueoglyxdbpdyqpgwgtmixlkxphtltijcm","maunpmlrxvreydzpzllevzlxpbkeeyrsskuht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String chars = "owqugdlpqrnasodvbemfhuzctbibeboxgdklfyzyucomprzzoxwwxm";
final int i = countCharacters(words, chars);
System.out.println("i = " + i);
}
/**
* 思路三：
* 93%
* 有以下注意的点：
* 1.使用字符串数组记录字符个数
* 2.使用flag变量控制是否包含该字符串（好像在哪里看到尽量不要使用label这种跳转的....）
* 3.如果字符串长度大于chars，可以不用比较，直接跳过
* 4.可以一边遍历目标字符串，一边比较，可能会提前结束，不用再用一个for循环。
*
* 作者：13217319563
* 链接：https://leetcode-cn.com/problems/find-words-that-can-be-formed-by-characters/solution/javazhi-xing-5mschao-guo-93-by-13217319563/
* 来源：力扣（LeetCode）
* 著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
* @param words
* @param chars
* @return
*/
public static int countCharacters(String[] words, String chars) {
int len = 0;
char[] cs = new char[26];
for (char c : chars.toCharArray()) {
cs[c - 'a']++;
}
for (String str : words) {
boolean flag = true;
//如果长度大于chars数组,直接不用比较
if (chars.length() < str.length()) {
flag = false;
} else {
char[] temp = new char[26];
for (char t : str.toCharArray()) {
temp[t - 'a']++;
if(temp[t - 'a'] > cs[t - 'a']) {
flag = false;
break;
}
}
}
if (flag) {
len += str.length();
}
}
return len;
}
/**
* 63.5%
* 思路二：
* 我们既统计“字母表”中字母出现的次数，也统计单词中字母出现的次数。
* 如果单词中每种字母出现的次数都小于等于字母表中字母出现的次数，那么这个单词就可以由字母表拼出来。
*
* 如何实现计数结构呢？一般的方法是用 map，比如 C++ 的 unordered_map 或者 Java 的 HashMap。
* 但是我们注意到题目有一个额外的条件：所有字符串中都仅包含小写英文字母。
* 这意味着我们可以用一个长度为 26 的数组来进行计数。这也是很多字符串计数问题的常用技巧。
*
* 作者：nettee
* 链接：https://leetcode-cn.com/problems/find-words-that-can-be-formed-by-characters/solution/tong-ji-zi-mu-chu-xian-de-ci-shu-shu-zu-ji-qiao-cj/
* 来源：力扣（LeetCode）
* 著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
* @param words
* @param chars
* @return
*/
// public static int countCharacters(String[] words, String chars) {
// int[] chars_count = count(chars); // 统计字母表的字母出现次数
// int res = 0;
// for (String word : words) {
// int[] word_count = count(word); // 统计单词的字母出现次数
// if (contains(chars_count, word_count)) {
// res += word.length();
// }
// }
// return res;
// }
//
// // 检查字母表的字母出现次数是否覆盖单词的字母出现次数
// static boolean contains(int[] chars_count, int[] word_count) {
// for (int i = 0; i < 26; i++) {
// if (chars_count[i] < word_count[i]) {
// return false;
// }
// }
// return true;
// }
//
// // 统计 26 个字母出现的次数
// static int[] count(String word) {
// int[] counter = new int[26];
// for (int i = 0; i < word.length(); i++) {
// char c = word.charAt(i);
// counter[c-'a']++;
// }
// return counter;
// }
/**
* 思路一：使用两个map集合存储word与chars的每个字符出现次数，然后比对，如果存在字符出现在word中，而不出现在chars中，
* 则返回0，如果存在word中的某字符数大于chars则返回0，否则返回word长度。
* @param words
* @param chars
* @return
*/
// public static int countCharacters(String[] words, String chars) {
// int i,num = 0;
// final Map getmap = getmap(chars);
// System.out.println("charsmap = " + getmap);
// for (String word :words){
// num += compare(word,getmap);
// }
// return num;
// }
// public static Map getmap(String str){
// int i;
// Map<String,Integer> charsmap = new HashMap();
// final String[] split = str.split("");
// for (i = 0;i<split.length;i++){
// if (charsmap.get(split[i]) == null){
// charsmap.put(split[i],1);
// }else {
// Integer integer = charsmap.get(split[i]);
// integer++;
// charsmap.put(split[i],integer);
// }
// }
// return charsmap;
// }
// public static int compare(String word, Map<String,Integer> chars){
// final Map<String,Integer> getmap = getmap(word);
// final Set set = getmap.keySet();
//
// int i;
// if (!chars.keySet().containsAll(set)){
// return 0;
// }else {
// final Iterator iterator = set.iterator();
// while (iterator.hasNext()){
// final Object next = iterator.next();
// if (getmap.get(next) > chars.get(next)){
// return 0;
// }
// }
// }
// return word.length();
// }
}
