如何使用递归查找相同数字的序列? - 爪哇
我有一个分配,例如,从用户那里获得了一个数字输入,例如:“ 57779227” 我需要返回最长的相同数字序列。在此示例中,最长的序列是“ 777”,返回应为3(因为数字“ 7”是连续的。
到目前为止,我编写了迭代方法。 ***在这种方法中不使用循环,只有递归。 ***
迭代示例:
public static int maxSequence(int num) {
int max = 1; //initiate
int currentCount = 1;
int prevDigit = 11;//Because num%10 != 11 Always!
int currentDigit;
while (num!=0) {
currentDigit = num%10;
if (prevDigit == currentDigit)
currentCount++;
else if (currentCount > max)
max = currentCount;
if (prevDigit != currentDigit) //initiate for the next Iteration
currentCount = 1;
prevDigit = currentDigit;
num = num/10;
}
return max;
}
I have an assignment where i get an number input from the user, for example : "57779227"
and i need to return the longest sequence of identical numbers. For this example, the longest sequence is "777" and the return should be 3 (as the amount of times the number "7" is in a row.
So far I wrote an iteration method.
***No loops to be used in this method, ONLY RECURSION. ***
Iteration example :
public static int maxSequence(int num) {
int max = 1; //initiate
int currentCount = 1;
int prevDigit = 11;//Because num%10 != 11 Always!
int currentDigit;
while (num!=0) {
currentDigit = num%10;
if (prevDigit == currentDigit)
currentCount++;
else if (currentCount > max)
max = currentCount;
if (prevDigit != currentDigit) //initiate for the next Iteration
currentCount = 1;
prevDigit = currentDigit;
num = num/10;
}
return max;
}
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当先前digit!= current -digit时,将开始一个新的计数
When previousDigit != currentDigit then a new count will be start