Learn how to check if two strings are anagrams in Java with step-by-step examples and code snippets.
此解法因为用到了map.containsKey()方法,所以时间复杂度最好的情况是O(n),最坏的情况是O(n^2),空间复杂度是O(n)。 publicbooleanisAnagram3(String s, String t){if(s ==null|| t ==null|| s.length() != t.length()) {returnfalse; } HashMap<Character,Integer> map =newHashMap<Character,I...
优化算法:通过选择合适的排序算法或字符比较方法,可以显著提高算法的性能。例如,使用计数排序(Counting Sort)或哈希排序(Hash Sort)可以显著提高排序速度。 缓存:利用缓存技术,例如LRU(Least Recently Used)缓存,存储最近使用的Anagram结果,从而减少计算时间。
https://leetcode.com/problems/find-all-anagrams-in-a-string/discuss/636988/Sliding-Window-or-HashTable-or-Java-Explained-with-Diagram-Beats-99 classSolution {publicList<Integer>findAnagrams(String s, String p) {int[] map =newint[26]; List<Integer> result =newArrayList<>();for(inti=0;i<...
Then store the count as value in the integer array Iterate and check for non zero count **/ class Solution { public boolean isAnagram(String s, String t) { Map<Character, Integer> charCountMap = new HashMap<>(); for(char c: s.toCharArray()) { charCountMap.put(c, charCountMap.get...
问给定字符串的Panagram解决方案EN实际上，您的解决方案相当不错，代码样式也很好，等等。我建议您做一些...
Java Program to check if two strings are an anagram or not import java.util.*; class test{ public static void main(String args[]){ String str1,str2; System.out.println("Enter the two String value"); Scanner sc=new Scanner(System.in); ...
意思是看两个string中各个字母出现次数是否相同,用数组存储就行(哈希),结果: Success Runtime: 8 ms, faster than 98.42% of C++ online submissions for Valid Anagram. Memory Usage: 9.4 MB, less than 50.10% of C++ online submissions for Valid Anagram. 代码: class Solution { public: bool isAnagram...
// Java implementation of the approach public class GFG { // Function to return the cost to make str a Panagram static int costToPanagram(String str, int cost[]) { int i, n = str.length(); int occurrences[] = new int[26]; // Count the occurrences of each lowercase character for...
#include<string.h> float val[13]; int num, res[13]; char ch[13]; void dfs(int cur) { if (cur == num) { for (int i = 0; i < num; i++) printf("%c", ch[res[i]]); printf("\n"); return; } for (int i = 0; i < num; i++) ...