说我有两个弦，
str1 = "abcdefg"
str2 = "cdefghijkl"
如果用户指定了整数，则说n = 4。
如何确定两个字符串是否重叠4个或更多字符？
在这种情况下，它们将“ cdefg”重叠5以使“ abcdefghijkl”
如果n = 6，则它们将不会重叠，因为“ bcdefg”！=“ cdefgh”
我遇到的问题是，如果重叠超过用户指定的数量。
最佳答案
我遇到的问题是，如果重叠超过用户指定的数量。
这就是循环的开始：从用户重叠定义的位置开始，并不断增加直到（1）检测到重叠或（2）字符串之一中的字符用完。
伪代码如下所示：
int userOverlap = ... // Get user overlap
int minLength = Math.min(stringOne.length(), stringTwo.length());
for (int overlap = userOverlap ; overlap <= minLength; overlap++) {
if (testOverlap(stringOne, stringTwo, overlap)) {
return true;
}
}
return false;
private static bool testOverlap(String a, String b, int overlap) {
... // This is your method that tests for one specific overlap
}