然后用两个变量left和right表示滑动窗口的左右边界,用变量cnt表示字符串p中需要匹配的字符个数,然后开始循环,如果右边界的字符已经在哈希表中了,说明该字符在p中有出现,则cnt自减1,然后哈希表中该字符个数自减1,右边界自加1,如果此时cnt减为0了,说明p中的字符都匹配上了,那么将此时左边界加入结果res中。
Find All Anagrams in a String 找变位子串: Given a strings and anon-empty stringp, find all the start indices ofp's anagrams ins. Strings consists of lowercase English letters only and the length of both stringss andp will not be larger than 20,100. The order of output does not matter....
UsefindMethod to Find Substring in a String in C++ The most straightforward way of solving the issue is thefindfunction, which is a built-instringmethod, and it takes anotherstringobject as an argument. #include<iostream>#include<string>using std::cin;using std::cout;using std::endl using ...
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2、string构造函数 void test01() { string s1;//默认构造 const char* str = "hello world"; string s2(str); cout << "s2=" << s2 << endl; string s3(s2); cout << "s3=" << s3 << endl; string s4(10, 'a');//10个a cout << "s4=" << s4 << endl; } 3、string赋值...
publicstaticvoidmain(String[]args) { Stringtext="AABCCAAADCBBAADBBC"; Stringstr="AA"; intcount=StringUtils.countMatches(text,str); System.out.println(count); } } Download Code Output: 3 That’s all about finding the occurrences of a substring in a string in Java. ...
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function findLongestWord(str) { // 分割字符串 var stringArr = str.split(" "); // 边界值及递归跳出条件 if (stringArr.length === 1) { return stringArr[0].length; } // 比较数组中第一个元素(字符串)与第二个元素(字符串)的长度 if (stringArr[0].length >= stringArr[1].length) {...
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Sub in字母get数字() ' Dim a As String a= InputBox(prompt:="请输入列字母") If a <> "" Then MsgBox Range("a1:" & a & "1").Count ‘取得这个范围的总列数就是我们要的列数字啦 Else MsgBox "你没输入" Exit Sub End If End Sub ...