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To find the first occurrence of x at or after position N, you can use s1.indexOf (x,N). s1.compareTo (s2) is an integer-valued function that compares the two strings. If the strings are equal, the value returned is zero. If s1 is less than s2, the value returned is -1, and if s1 is greater than s2, the value returned is 1.
The next diagram illustrates the resulting situation. If we write the expression sl == s2, Java evaluates this by comparing the values stored in s1 and s2, not the strings to which s1 and s2 refer.
String s1 = "Hello"; String s2 = new String("Hello"); System.out.println(s1 == s2); This returns: false This is because the == operator doesn't check for equality. It checks for identity. In other words, it doesn't compare the String s value - it compares object references. The s1 is a reference variable to the same object in memory that s2 references. This is because the String Pool doesn't ...
Given two strings S1 and S2, the task is to check if it's possible to generate string S2 by repeatedly appending subsequences of S1 to the end of an initially empty string.
这是以前刚学Java那会经常被问到的问题，并且也会经常出现在面试中。 这个问题不仅仅只局限于Java，可以适用于其他大部分语言，正好最近有个小朋友在学编程，特此整理一下。 String s1 = "abc" ;String s2 = new String ("abc");首先，我们先看看以上两句分别做了什么？
先來看 case 1， s1 + s2 運算可以讓兩個 string 去產生一個新個 string，然後複製回去 s1，可想而知很沒效率。 case 2 與 case 3 例子中，概念都是把 s2 加到原本 s1 的後面，所以 case 2 和 case 3 實際執行效率是一樣的。
因为equals方法比较的真正的char数据，而s1和s2最终指向的都是同一个char数组对象，所以s1.equals (s2)等于true。 关于他们最终指向的都是同一个char数组对象这一观点，也可以通过反射证明：
string字符串的基本使用string 类型的操作函数 s.empty () 如果s为空串，则返回true,否则返回false s.size () 返回s中字符的个数，s.length ()用法与s.size ()相同 s [n] 返回s中位置为n的字符，位置从0开始计数 (可将值赋给s [n]) s1+s2 把s1和s2连接成一个新字符串，返回新生成的字符串 s1=s2 把s1内容替换为s2的副本 v1 ...
Since we used string literals this time rather than the new keyword, the Java run-time will check if that string literal already exists as an object in memory, and if so reuse it. So s1 and s2 will refer to the same string object. That means that both == and equals will be true.
也就是说str2指向的String对象并没有驻留到字符串常量池，而str1指向的对象驻留到字符串常量池里面去了，且他们并不是同一个对象。