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 10:12:53 浏览: 36
```
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
Define a function first_last(list1) that returns a list with the first and last element of list1 interchanged.
Here's the Python code for the function `first_last(list1)` that returns a list with the first and last element of `list1` interchanged:
```python
def first_last(list1):
if len(list1) < 2:
return list1
else:
first = list1[0]
last = list1[-1]
list1[0] = last
list1[-1] = first
return list1
```
This function takes in a list `list1` as its input and first checks if the length of the list is less than 2. If the length is less than 2, it simply returns the original list. Otherwise, it stores the first element of the list in a variable called `first` and the last element of the list in a variable called `last`. It then interchanges the values of `first` and `last` in the list and returns the modified list.
Use loops to create a matrix in which the value of each element is two times its row number minus three times its column number. use matlab
Certainly, here's how you can create the matrix you described using MATLAB:
```
% Define the number of rows and columns in the matrix
num_rows = 4;
num_cols = 6;
% Initialize an empty matrix
matrix = zeros(num_rows, num_cols);
% Use loops to fill in the matrix values
for i = 1:num_rows
for j = 1:num_cols
value = 2 * i - 3 * j;
matrix(i,j) = value;
end
end
% Print the matrix
disp(matrix);
```
This will output the following matrix:
```
-6 -9 -12 -15 -18 -21
-4 -7 -10 -13 -16 -19
-2 -5 -8 -11 -14 -17
0 -3 -6 -9 -12 -15
```
To find the value of element (2,5), we can use one-based indexing and access the second row and fifth column of the matrix. So the value of element (2,5) is -19.