帮我完成以下代码。We define a runs in a list is a series of 2 or more adjacent elements of the same value. Write a method to return the number of runs in the input list. Test case 1: countRuns([1, 2, 2, 2, 3]) = 1 (which is 2, 2, 2) Test case 2: countRuns([1, 1, 2, 3, 4, 5, 5]) = 2 (which is 1, 1; and 5, 5). public static int countRuns(List<Integer> list) {}
时间: 2023-07-14 07:12:53 浏览: 66
```
public static int countRuns(List<Integer> list) {
int count = 0;
int currentRun = 1;
for (int i = 1; i < list.size(); i++) {
if (list.get(i).equals(list.get(i-1))) {
currentRun++;
} else {
if (currentRun >= 2) {
count++;
}
currentRun = 1; // reset current run
}
}
if (currentRun >= 2) {
count++;
}
return count;
}
```
这个方法遍历列表,计算连续相同元素的长度。如果这个长度大于等于2,就将计数器加1。最后返回计数器的值。
相关问题
We will write a procedure every-nth that accepts two arguments: num is an integer and list-of-sents is a list of a certain number of lists. every-nth selects the num-th elements from each list in list-of-sents and returns the list of selected elements. Assume that all lists in listof-sents have more than num elements. Consider the first element as the 0-th. Examples: (every-nth 2 '((a b c d) (e f g h))) gives (c g). (every-nth 1 '((a b c) (d e f) (g h i))) gives (b e h). (define (every-nth num list-of-sents) ; You must fill in this part )
Here's one way to implement the `every-nth` procedure:
```
(define (every-nth num list-of-sents)
(apply map
(lambda args (list-ref args num))
list-of-sents))
```
In this implementation, we use `map` to apply a function to each element of `list-of-sents`. The function we apply is a lambda that takes `num` arguments (one from each sublist) and returns the `num`-th element of the sublist. We use `apply` to pass the sublists as separate arguments to `map`.
Note that we assume that all sublists have at least `num+1` elements, since we count the first element as the 0-th. If some sublists might have fewer elements, we should add a check to avoid an index out of bounds error.
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