Effective alphabetic words

Organizing thoughts: I am mentally retarded now, so no matter how simple the topic is, I will share it. My idea is to define two dictionaries (such a simple topic, unexpectedly thought of using dictionaries, wonderful brain circuit). Train to traverse two strings and add the value of corresponding key + 1. Finally compare the values of two dictionaries.

import java.util.HashMap;
import java.util.Map;

public class num_11 {

	public static void main(String[] args) {
		String s = "hhbywxfzydbppjxnbhezsxepfexkzofxyqdvcgdvgnjbvihgufvgtuxtpioxgjuwawkbaerbsirrktcjcesepcocaglbassivfbjhikynfsvlbtkawstrclbkpuldshfgfvwjawownulsggsxhhqglbhjlgltfrqyjntgldlgorgatnjrsywlyngxrcwfojkmydmjbzuquqnzbpwlcbdzgdicctpkgtbdjgywkyrkityvohjbuvmzdfyicyplmrpygdhaeqbuxnehfybsysnnmzbhprsyjmtpcrzeqqbirofzjtlsyofdyeffkvpuhzaflwjfhnsuyryetjuajjnjwvlvrhvpenaarnzoafztixjrisfzdlepcwhxudjpsiirtofymnovacjmpdjtethjqfwduekczlqhsfjgqesyoxcfooagrdhyvsmssbhsclnlblhobvhwtpyftolneozlhbtjagpgqnnapktyevdvjfwdnbwsbelweoflhyhifprieuvfcdkavqxkygjlaegqfmzndgxbsccjgpclxmlpstrqjtqyvlqcopnahqvnpvkjimfdlosvletmamqjvotqwhadutmfvlgldniixvdkmymfadckuaglgbuttymoqmzkaeqxugsrnfyxzcamwxujgzupefretsvbdweuwwcizjvhcowtmwgkdafcpzctpsjwdocgofivyshwdinbdhbxdfhjsrrsfchxkeqndgzauyprwfnrbmunanqnhmjhrufoinakwaciaoerioqffmipfqujfxwofqdyjbhagkyvmnxcwomgnmwlaodxgkgthnuctoozxrebjiynjwohtgukyneyofabpfdrxklopmxxiwjuxqpaazknscagfiaetmmwnwpzceglupqvlctywtpluoqbzdultcsudubqclbwlxyfboimfwriugfulbntvdwnxgiycxvennakpodorvpiknkridcumsovvfzikiydqewgjhacrkxddqpncirzlsynbjwutomcwphgggcsnfqxwxguokjvihoewffngivnkapaqjrsshqfbpfqhzihzndzeyznzyndmhfmypgwsqyernninzijwrirmbebsupvukfjddsjbebfedtdcufnubxqocqeqaincaadorteimkecpkdrjpprqtrllypfiktvcfadfotgjghhtkctgidwreyjulbtmtpwicbqalejtxsgjqepzeduvwvjbbxaapwvmjuweurhdbacxaygzdyitwcwlguknhqxdrfpfhstgnwmwagcblxkqjjwxfhmpunstfhkiffqchcohgcuktfplffqtwlwcsddkaeijajkdpmexteafezquyfonocuwuxrvigpcqbzmxtrvctovnvrobyjmjnqdhxqjxtltichttiuemvxixfeyqzuzxkwmyclbwutbvrgtefcdizjnkmlsvrujtiqhrwpnzkdtkwogypvpittpfhjtlrjtilubngkkjochucmdudfcpudjhiwaflmdfgsjqlwqxlgmdwouofjnytwxumjaailltmzrrfvqcawkzytomjaaafknfnaaylmqkvyeljnznomervgdhnqfhtpalziumhilupdsjpvzwjsgqtnovjneotcsfdtqqtfwjlhsbhumoyjnpuniqeszknvzitifqmiezbzvtzoswezulpfylopuecqcefueesjfhvyddsvozowqtyphlnuridygcmjhjtfoenwbgtramegbnabtooohogyoqfmskrpayzbrsidfbqextanotbkzopgdidswzxveqtojspsbwdqsrrspsyddgwntndfqfqeuvmrnyjvdrpszghetjnbpitorzonhihbzncpwvnwndriupumwihnpxskwsvjqzuetprscqnihpfebqobdezwohlubpyoulnkoicmtjyfsmvjltsxtvydxkajhvkcmhrkeuudrgxjezoeessffmheelwvotkjecsnwybtqutmkubtzbwshbacffkodyhgv";
		String t = "hhbywxfzydbppjxnbhezsxepfexkzofxyqdvcgdvgnjbvihgufvgtuxtpioxgjuwawkbaerbsirrktcjcesepcocaglbassivfbjhikynfsvlbtkawstrclbkpuldshfgfvwjawownulsggsxhhqglbhjlgltfrqyjntgldlgorgatnjrsywlyngxrcwfojkmydmjbzuquqnzbpwlcbdzgdicctpkgtbdjgywkyrkityvohjbuvmzdfyicyplmrpygdhaeqbuxnehfybsysnnmzbhprsyjmtpcrzeqqbirofzjtlsyofdyeffkvpuhzaflwjfhnsuyryetjuajjnjwvlvrhvpenaarnzoafztixjrisfzdlepcwhxudjpsiirtofymnovacjmpdjtethjqfwduekczlqhsfjgqesyoxcfooagrdhyvsmssbhsclnlblhobvhwtpyftolneozlhbtjagpgqnnapktyevdvjfwdnbwsbelweoflhyhifprieuvfcdkavqxkygjlaegqfmzndgxbsccjgpclxmlpstrqjtqyvlqcopnahqvnpvkjimfdlosvletmamqjvotqwhadutmfvlgldniixvdkmymfadckuaglgbuttymoqmzkaeqxugsrnfyxzcamwxujgzupefretsvbdweuwwcizjvhcowtmwgkdafcpzctpsjwdocgofivyshwdinbdhbxdfhjsrrsfchxkeqndgzauyprwfnrbmunanqnhmjhrufoinakwaciaoerioqffmipfqujfxwofqdyjbhagkyvmnxcwomgnmwlaodxgkgthnuctoozxrebjiynjwohtgukyneyofabpfdrxklopmxxiwjuxqpaazknscagfiaetmmwnwpzceglupqvlctywtpluoqbzdultcsudubqclbwlxyfboimfwriugfulbntvdwnxgiycxvennakpodorvpiknkridcumsovvfzikiydqewgjhacrkxddqpncirzlsynbjwutomcwphgggcsnfqxwxguokjvihoewffngivnkapaqjrsshqfbpfqhzihzndzeyznzyndmhfmypgwsqyernninzijwrirmbebsupvukfjddsjbebfedtdcufnubxqocqeqaincaadorteimkecpkdrjpprqtrllypfiktvcfadfotgjghhtkctgidwreyjulbtmtpwicbqalejtxsgjqepzeduvwvjbbxaapwvmjuweurhdbacxaygzdyitwcwlguknhqxdrfpfhstgnwmwagcblxkqjjwxfhmpunstfhkiffqchcohgcuktfplffqtwlwcsddkaeijajkdpmexteafezquyfonocuwuxrvigpcqbzmxtrvctovnvrobyjmjnqdhxqjxtltichttiuemvxixfeyqzuzxkwmyclbwutbvrgtefcdizjnkmlsvrujtiqhrwpnzkdtkwogypvpittpfhjtlrjtilubngkkjochucmdudfcpudjhiwaflmdfgsjqlwqxlgmdwouofjnytwxumjaailltmzrrfvqcawkzytomjaaafknfnaaylmqkvyeljnznomervgdhnqfhtpalziumhilupdsjpvzwjsgqtnovjneotcsfdtqqtfwjlhsbhumoyjnpuniqeszknvzitifqmiezbzvtzoswezulpfylopuecqcefueesjfhvyddsvozowqtyphlnuridygcmjhjtfoenwbgtramegbnabtooohogyoqfmskrpayzbrsidfbqextanotbkzopgdidswzxveqtojspsbwdqsrrspsyddgwntndfqfqeuvmrnyjvdrpszghetjnbpitorzonhihbzncpwvnwndriupumwihnpxskwsvjqzuetprscqnihpfebqobdezwohlubpyoulnkoicmtjyfsmvjltsxtvydxkajhvkcmhrkeuudrgxjezoeessffmheelwvotkjecsnwybtqutmkubtzbwshbacffkodyhgv";
		System.out.println(isAnagram(s, t));

	}

	public static boolean isAnagram(String s, String t) {
		Map<Character, Integer> map1 = new HashMap<Character, Integer>();
		Map<Character, Integer> map2 = new HashMap<Character, Integer>();
		for (int i = 0; i < s.length(); i++) {
			Integer val = map1.get(s.charAt(i));
			map1.put(s.charAt(i), (val == null ? 1 : ++val));
		}
		for (int i = 0; i < t.length(); i++) {
			Integer val = map2.get(t.charAt(i));
			map2.put(t.charAt(i), (val == null ? 1 : ++val));
		}

		System.out.println(map1.values());
		System.out.println(map2);
		if (map1.size() != map2.size())
			return false;

		for (Character c : map1.keySet()) {
			if ((map1.get(c) != map2.get(c)) || !map2.containsKey(c)) {
				return false;
			}
		}

		return true;
	}

}

Above is my code, put it into leetcode, it can't run well, but it's OK to test it on my computer. I hope you can correct me.

In the process of doing this problem, I saw a big blogger, deeply impressed!!

https://blog.csdn.net/lu_wei_keke/article/details/80975608

You can have a look!

Tags: Java

Posted on Mon, 02 Dec 2019 09:39:32 -0500 by nosmasu