【每日一题】Day0012:力扣题库NO.1044. 最长重复子串

今天的题虽然通过自己的思路应该是能够解出答案，但是提交超时；去学习别人的思路和解法，因为没有算法基础也难懂，不过姑且先将自己的思路述出。

题目链接：

力扣1044最长重复子串https://leetcode-cn.com/problems/longest-duplicate-substring/思路很简单，都在注释里直接看代码：

package cn.daycode.leetcode;

public class LongestDupSubstring {
    public static void main(String[] args) {
        Solution s = new Solution();

        // 测试超时的用例    
        System.out.println(s.longestDupSubstring("unuoqslvgaekunapeaapsxbodakcrapmmfuxoowlysenyiygzhawbfbuxlswtxzzhlspfgzckzuoxizijrluvndcpnknsvrnfulksqqxdznfjgjlocozuolroealyhhwrzcsuzgcsekkqmwlnylsscpmynjchwvpwebwkraudfqboeqmxhghekcstrcozldbwyniwqmyrbkzwvknulehzxwckwohlmjvlqyexvzzlptbnnbctywyezcssubmiukcsixmnapgttmlfwtwnmxqgsobqsnxucwrszxcuprnlxjteytwqzridgtgkkbsekeytuxslavhebbpjewdwlhchtzwohqkojiqneahxacutymteoyloottnsmmrphngclortfvudoeckxkjqatxqmvboumxmdrxoxpyprkwvfpviqkbhsjdeosxdbzzfsomrxojkokofrrdoavhjufyayisibcgngqprircpcjyvynhzmrtyynfgfgeomscywbcvmlhcmyesuxwurmpdxhddyfnumgkvphmtqzrbsbjeliyrqfswrchkurvqtsmzchjfvotdmueabpllagtfvefssmgevrzydftuvxqdbdrpysifoxgvlkljhkiksoxfndxayrsaxeoefmimnqfeerggwxsbyarfvfwvrmtwjainqposafhwysxaaegzeewhyrsfnduqihpwipeewnubscqpvjekikyiwmpwynydbnvylqydgiwsenvulkknlbuqpwhxqclrnuvdwqpxkksyewsklffwtggbmxnttvnjbiqjdoezffmmighfdcirjakacncwckjqwqsjxsbsxbaeaagmnybxpwvucyskfkydlkxdlfmodygcxzimmhdgmwwqxrflldkwvyqlfwasxxoetlgtopmqnfkfvkshkrikfcvdrguxciupphxyssowakrnsiutntlliedxnqkldzfkqefgvyaargpqbknlmlbejkuupolrhapowrgciyteeeulpldvlqkklwsxtvcmxjozbfqgmslutnkspfoaeylxufmuropnafrjveufqymddcrzvbdkhghzvbjzxyuibkemslbsluzwnduiutluergimcjsziusnqhyazuhskekblpoexmurndswlwxwieujdyilfntaaphwvnauvwmihlfphfpxzkwinxitiqxamjqxtdxswgxfmsfuqwtqqnzracsoerlrntgyypdroiuqtiawxeogqcrxxbjmgkqrlrbgeglkamqfvgaiipuxcvllcujnnlfzidrvzowaupaucbxzwzhueqqdphpmsdszdtcycrmiwyhslbvknlxhfgpcbotvarakaymjgfydzfmnxwausfhdxulmxhvzgbswfdwrtitwyibwfnpsekmymsglzpiyrhnuaatmqkjqhcwycnaxjbxqgprgoesjnupyfhbyounmbwgjfplslyudxyaxuspyxjcuorzxcryiuwlgprpurjdxozsalkbjcjatnxstbxinrokiufjldyuyngkiobomuemplhziwaapevhrluxohodizmmqznpbggcaqfgobgpvytmbpdmrrcbzbgrppmlmigghuzrdfhwhksgmmnjnbglrrckwbdstsjdjxaeopxawedvszchktinhceceauookzcwbdkbglzkedrjuqhrnaxluttcjbbdmqimlgppvuvmjqipkcjvyzhingwymthqrpguhvlcjqrstczkdvdrwwtwdrakmbxzjfsszpziurglmpsyllgxtbewoftnhvzmtygsvbnwhltwqszqdopzxbwyrdyrpffnilfdxtlofyicjshyrpncsqndubqkgkgrtbybulwegugcvfqhpfnvvccocolfpiyojgoranoyegkpitmfkmdkyihdqscwykyhgkyljypjobeijjvdpjglbkefxtsmyyytluqpvdcsgrtvsamzupdjyzqtcvpoquialxzaeerxtqkitsapniklylrztgclbxyzzrkanhucwdccdvafyirulajxbmeyaoqtqpprkohdddzdxtiwwctcevneriqahvrgyqvyrwfcvvqbhxmiajqkwhztepimveqisonqccfgwnhpkkqsqlmqqszwjxcpmyzldhjmriayuxpfosdtmhvtncuboiggyroxgwlbybievmdwaverpetzrmeogojuyipkwsibihuhzxjnwugjpwonfsbwaxfcqveyipwkhkzsfarvxrdlxvoiduextlcnnwuaiishdbfnxvfzkqfbqynfznbqijfkamnhlwcmkysbotlebtyfkoptfignbwxfldmbebmtyoggfvbyyygbibfhgfrdtwptwihjndlxmtdkntxtfjwstbmukwlwkzecogozerskubaxjqtliwpgyfwvkqqagmwmnfvuoujlelopdbfqyjnemaibfmgfhqzoptcolufrgafmwkzlxaudgvkhicwjdyomiqgoapvhyakslmhnznivjulyyqusxolkmfyskcfgshotfdjbtmflvtagraeazawnbtcquiivziijjporrsuytdkayjtnexvwtoswqmncbvwsxcaowrjmqdtyzqdjrqbmnaqqlkbdsrweeofeztqlgijzsriayrlknxcinibqqguxztuikzetcwipuchwukczxunuoqslvgaekunapeaapsxbodakcrapmmfuxoowlysenyiygzhawbfbuxlswtxzzhlspfgzckzuoxizijrluvndcpnknsvrnfulksqqxdznfjgjlocozuolroealyhhwrzcsuzgcsekkqmwlnylsscpmynjchwvpwebwkraudfqboeqmxhghekcstrcozldbwyniwqmyrbkzwvknulehzxwckwohlmjvlqyexvzzlptbnnbctywyezcssubmiukcsixmnapgttmlfwtwnmxqgsobqsnxucwrszxcuprnlxjteytwqzridgtgkkbsekeytuxslavhebbpjewdwlhchtzwohqkojiqneahxacutymteoyloottnsmmrphngclortfvudoeckxkjqatxqmvboumxmdrxoxpyprkwvfpviqkbhsjdeosxdbzzfsomrxojkokofrrdoavhjufyayisibcgngqprircpcjyvynhzmrtyynfgfgeomscywbcvmlhcmyesuxwurmpdxhddyfnumgkvphmtqzrbsbjeliyrqfswrchkurvqtsmzchjfvotdmueabpllagtfvefssmgevrzydftuvxqdbdrpysifoxgvlkljhkiksoxfndxayrsaxeoefmimnqfeerggwxsbyarfvfwvrmtwjainqposafhwysxaaegzeewhyrsfnduqihpwipeewnubscqpvjekikyiwmpwynydbnvylqydgiwsenvulkknlbuqpwhxqclrnuvdwqpxkksyewsklffwtggbmxnttvnjbiqjdoezffmmighfdcirjakacncwckjqwqsjxsbsxbaeaagmnybxpwvucyskfkydlkxdlfmodygcxzimmhdgmwwqxrflldkwvyqlfwasxxoetlgtopmqnfkfvkshkrikfcvdrguxciupphxyssowakrnsiutntlliedxnqkldzfkqefgvyaargpqbknlmlbejkuupolrhapowrgciyteeeulpldvlqkklwsxtvcmxjozbfqgmslutnkspfoaeylxufmuropnafrjveufqymddcrzvbdkhghzvbjzxyuibkemslbsluzwnduiutluergimcjsziusnqhyazuhskekblpoexmurndswlwxwieujdyilfntaaphwvnauvwmihlfphfpxzkwinxitiqxamjqxtdxswgxfmsfuqwtqqnzracsoerlrntgyypdroiuqtiawxeogqcrxxbjmgkqrlrbgeglkamqfvgaiipuxcvllcujnnlfzidrvzowaupaucbxzwzhueqqdphpmsdszdtcycrmiwyhslbvknlxhfgpcbotvarakaymjgfydzfmnxwausfhdxulmxhvzgbswfdwrtitwyibwfnpsekmymsglzpiyrhnuaatmqkjqhcwycnaxjbxqgprgoesjnupyfhbyounmbwgjfplslyudxyaxuspyxjcuorzxcryiuwlgprpurjdxozsalkbjcjatnxstbxinrokiufjldyuyngkiobomuemplhziwaapevhrluxohodizmmqznpbggcaqfgobgpvytmbpdmrrcbzbgrppmlmigghuzrdfhwhksgmmnjnbglrrckwbdstsjdjxaeopxawedvszchktinhceceauookzcwbdkbglzkedrjuqhrnaxluttcjbbdmqimlgppvuvmjqipkcjvyzhingwymthqrpguhvlcjqrstczkdvdrwwtwdrakmbxzjfsszpziurglmpsyllgxtbewoftnhvzmtygsvbnwhltwqszqdopzxbwyrdyrpffnilfdxtlofyicjshyrpncsqndubqkgkgrtbybulwegugcvfqhpfnvvccocolfpiyojgoranoyegkpitmfkmdkyihdqscwykyhgkyljypjobeijjvdpjglbkefxtsmyyytluqpvdcsgrtvsamzupdjyzqtcvpoquialxzaeerxtqkitsapniklylrztgclbxyzzrkanhucwdccdvafyirulajxbmeyaoqtqpprkohdddzdxtiwwctcevneriqahvrgyqvyrwfcvvqbhxmiajqkwhztepimveqisonqccfgwnhpkkqsqlmqqszwjxcpmyzldhjmriayuxpfosdtmhvtncuboiggyroxgwlbybievmdwaverpetzrmeogojuyipkwsibihuhzxjnwugjpwonfsbwaxfcqveyipwkhkzsfarvxrdlxvoiduextlcnnwuaiishdbfnxvfzkqfbqynfznbqijfkamnhlwcmkysbotlebtyfkoptfignbwxfldmbebmtyoggfvbyyygbibfhgfrdtwptwihjndlxmtdkntxtfjwstbmukwlwkzecogozerskubaxjqtliwpgyfwvkqqagmwmnfvuoujlelopdbfqyjnemaibfmgfhqzoptcolufrgafmwkzlxaudgvkhicwjdyomiqgoapvhyakslmhnznivjulyyqusxolkmfyskcfgshotfdjbtmflvtagraeazawnbtcquiivziijjporrsuytdkayjtnexvwtoswqmncbvwsxcaowrjmqdtyzqdjrqbmnaqqlkbdsrweeofeztqlgijzsriayrlknxcinibqqguxztuikzetcwipuchwukczx"));
    }
}

class Solution {
    public String longestDupSubstring(String s) {
        String str = "";
        int tempLength = 0;
        int maxLength = 0;

        // 从头开始遍历
        for (int i = 0; i < s.length()-1; i++) {
            tempLength = 0;
            // 找到从当前下标开始往后最长的连续子串
            // 不是整个字符串（是以当前下标开始到结尾的子串）
            while(i+tempLength < s.length() && s.substring(i+1).contains(s.substring(i, i+1+tempLength))){
                tempLength++;
            }
            // 如果找到的子串长度比当前已知的最长子串长，就记录一下该字符串和长度
            if(tempLength > maxLength){
                str = s.substring(i,i+tempLength);
                maxLength = tempLength;
            }
        }

        return str;
    }

}

这题我觉得需要mark一下，之后继续研究一下。把大佬的思路和所用的算法搞清楚，自己能够独立复现。