[特殊字符] 如何判断两个字符串是否完全相同？三种方法带你搞定！

给定两个字符串判断它们是否完全相同区分大小写。示例输入: s1 “abc”, s2 “abc” → 输出: Yes输入: s1 “”, s2 “” → 输出: Yes输入: s1 “GeeksforGeeks”, s2 “Geeks” → 输出: No目录方法一直接使用运算符、equals、方法二使用字符串比较函数方法三自己实现字符逐位比较方法一直接使用运算符、equals、思路直接使用语言的相等运算符或方法equals对两个字符串进行比较逐字符判断是否相同。C#includeiostreamusingnamespacestd;boolareStringsSame(string s1,string s2){returns1s2;}intmain(){string s1abc;string s2abcd;if(areStringsSame(s1,s2)){coutYesendl;}else{coutNoendl;}return0;}C#includestdio.h#includestring.h#includestdbool.hboolareStringsSame(constchar*s1,constchar*s2){returnstrcmp(s1,s2)0;}intmain(){constchar*s1abc;constchar*s2abcd;if(areStringsSame(s1,s2)){printf(Yes\n);}else{printf(No\n);}return0;}JavapublicclassGfG{publicstaticbooleanareStringsSame(Strings1,Strings2){returns1.equals(s2);}publicstaticvoidmain(String[]args){Strings1abc;Strings2abcd;if(areStringsSame(s1,s2)){System.out.println(Yes);}else{System.out.println(No);}}}PythondefareStringsSame(s1,s2):returns1s2if__name____main__:s1abcs2abcdifareStringsSame(s1,s2):print(Yes)else:print(No)C#usingSystem;classGfG{staticboolareStringsSame(strings1,strings2){returns1s2;}staticvoidMain(){strings1abc;strings2abcd;if(areStringsSame(s1,s2)){Console.WriteLine(Yes);}else{Console.WriteLine(No);}}}JavaScriptfunctionareStringsSame(s1,s2){returns1s2;}consts1abc;consts2abcd;if(areStringsSame(s1,s2)){console.log(Yes);}else{console.log(No);}输出No方法二使用字符串比较函数思路利用语言的字符串比较函数如 strcmp、equals、localeCompare 等这些函数会逐字符比较返回 0 表示相同非 0 表示不同。C#includeiostreamusingnamespacestd;boolareStringsSame(chars1[],chars2[]){returnstrcmp(s1,s2)0;}intmain(){chars1[]hello;chars2[]hello;if(areStringsSame(s1,s2)){coutYesendl;}else{coutNoendl;}return0;}学完字符串比较的几种方法是不是还想看看这些算法在内存里到底是怎么一步步跑起来的别急最近挖到的宝藏网站【图码】(https://totuma.cn) 真心推荐给准备408考研或数据结构期末考试的同学。它把60多种算法做成交互式动画不仅支持输入自定义数据生成可视化还能直接上传C/C/Java/Python代码自动解析遇到看不懂的地方随时点一下7×24小时的AI代码解释随叫随到。复习时用这个算法可视化工具配合全书级知识点理解效率直接翻倍。赶紧去试试吧图码-数据结构与算法交互式可视化平台访问网站https://totuma.cnC#includestdio.h#includestring.h#includestdbool.hboolareStringsSame(chars1[],chars2[]){returnstrcmp(s1,s2)0;}intmain(){chars1[]hello;chars2[]hello;if(areStringsSame(s1,s2)){printf(Yes\n);}else{printf(No\n);}return0;}Javaimportjava.util.Arrays;classGfG{publicstaticbooleanareStringsSame(Strings1,Strings2){returns1.equals(s2);}publicstaticvoidmain(String[]args){Strings1hello;Strings2hello;if(areStringsSame(s1,s2)){System.out.println(Yes);}else{System.out.println(No);}}}PythondefareStringsSame(s1,s2):returns1s2defmain():s1hellos2helloifareStringsSame(s1,s2):print(Yes)else:print(No)if__name____main__:main()C#usingSystem;classGfG{staticboolareStringsSame(strings1,strings2){returns1.Equals(s2);}staticvoidMain(){strings1hello;strings2hello;if(areStringsSame(s1,s2)){Console.WriteLine(Yes);}else{Console.WriteLine(No);}}}JavaScriptfunctionareStringsSame(s1,s2){returns1.localeCompare(s2)0;}lets1hello;lets2hello;if(areStringsSame(s1,s2)){console.log(Yes);}else{console.log(No);}输出Yes方法三自己实现字符逐位比较思路先比较两个字符串的长度如果长度不同则直接返回 false。然后从第一个字符开始逐个比较每个字符直到发现不同或遍历结束。C#includeiostreamusingnamespacestd;boolareStringsEqual(strings1,strings2){if(s1.length()!s2.length()){returnfalse;}for(size_t i0;is1.length();i){if(s1[i]!s2[i]){returnfalse;}}returntrue;}intmain(){string s1hello;string s2hello;if(areStringsEqual(s1,s2)){coutYesendl;}else{coutNoendl;}return0;}C#includestdio.h#includestring.h#includestdbool.hboolareStringsEqual(constchar*s1,constchar*s2){if(strlen(s1)!strlen(s2)){return0;}for(size_ti0;istrlen(s1);i){if(s1[i]!s2[i]){return0;}}return1;}intmain(){constchar*s1hello;constchar*s2hello;if(areStringsEqual(s1,s2)){printf(Yes\n);}else{printf(No\n);}return0;}JavapublicclassGfG{publicstaticbooleanareStringsEqual(Strings1,Strings2){if(s1.length()!s2.length()){returnfalse;}for(inti0;is1.length();i){if(s1.charAt(i)!s2.charAt(i)){returnfalse;}}returntrue;}publicstaticvoidmain(String[]args){Strings1hello;Strings2hello;if(areStringsEqual(s1,s2)){System.out.println(Yes);}else{System.out.println(No);}}}Pythondefare_strings_equal(s1,s2):iflen(s1)!len(s2):returnFalseforiinrange(len(s1)):ifs1[i]!s2[i]:returnFalsereturnTruedefmain():s1hellos2helloifare_strings_equal(s1,s2):print(Yes)else:print(No)if__name____main__:main()JavaScriptfunctionareStringsEqual(s1,s2){if(s1.length!s2.length){returnfalse;}for(leti0;is1.length;i){if(s1[i]!s2[i]){returnfalse;}}returntrue;}consts1hello;consts2hello;if(areStringsEqual(s1,s2)){console.log(Yes);}else{console.log(No);}输出Yes总结方法时间复杂度空间复杂度说明直接使用运算符O(n)O(1)最简单推荐日常使用使用比较函数O(n)O(1)适合 C 语言等自己实现O(n)O(1)适合学习底层原理无论哪种方法核心思想都是逐字符比较注意区分大小写哦