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String s1,s2; (1 )s1=s1+s2;(2)s1+=s2; (1)和(2)有什么区别吗?+=有...
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String s1="abc"; //pool String s2="abccde"; //pool String s3="cde"; //pool String s4=s1+s3; //heap String s5=new String("abccde"); //heap System.out.println(s2==s4 ...
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 Strings 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 ...
Swap 'a' and 'c' in S1 and the resultant string is equal to S2. Input: S1 = "abcd", S2 = "abcdcd" Output: No . Approach: Create a string even_s1 from the characters at even indices from S1. Similarly, generate the strings even_s2, odd_s1 and odd_s2. Sort all the four strings from the previous steps.
Rearrange the string S1 as "dcba". Now, string S2 = "ab" is not a subsequence of S1. Input: S1 = "aa", S2 = "a" Output:-1. Approach: Follow the steps below to solve the problem: Store the frequency of characters in string S2 in an auxiliary array, say cnt[26]. Initialize a variable, say ch, to store the unique characters present ...
s1.indexOf(s2) returns an integer. If s2 occurs as a substring of s1, then the returned value is the starting position of that substring. Otherwise, the returned value is -1. You can also use s1.indexOf(ch) to search for a particular character, ch, in s1. To find the first occurrence of x at or after position N, you can use s1.indexOf(x,N). s1 ...
If two string variables, s1 and s2 reference the same portion of memory, they certainly represent the same string and "s1 ==s 2" will return true. However, it is also possible to have s1 and s2 reference two identical strings in different portions of memory, in which case "s1 ==s2" will return false.
s3 == s1. Reason — Each of the four comparisons are explained below: The first comparison uses the == operator to compare s1 and s2.== operator checks for reference equality, i.e., whether both variables refer to the same object in memory. Since s1 and s2 have the same value and were created using the same string literal, they will refer to the same object in memory.
Input 1. S1: "ccbdbd" S2: "cbd" Output 1 . True. Reason. We will first add S2 to an empty string. We now have "cbd." We will again add S2 after the 'c' character, i.e., after the 0th index. Now, we have "ccbdbd," which is the same as S1 and is formed by repeated insertions of String S2 into an empty string. Let's take another example. Input 2
In other words, one of the first string's permutations is the substring of the second string. Example 1: Input: s1 = "ab" s2 = "eidbaooo" Output: True Explanation: ...
The resulting string is s1 = "cbadba". - Choose the indices i = 2, j = 4. The resulting string is s1 = "cbbdaa". - Choose the indices i = 1, j = 5. The resulting string is s1 = "cabdab" = s2. Example 2: Input: s1 = "abe", s2 = "bea" Output: false Explanation: It is not possible to make the two strings equal. Constraints: * n == s1.length == s2 ...