Given a string, return the length of the largest "block" in the string. A block is a run of adjacent chars that are the same.

时间: 2024-05-10 19:15:59 浏览: 5
Sure, I can help you with that. Here's a Python function that takes a string as input and returns the length of the largest block: ```python def largest_block(s): max_block_len = 0 block_len = 1 for i in range(1, len(s)): if s[i] == s[i-1]: block_len += 1 else: max_block_len = max(max_block_len, block_len) block_len = 1 return max(max_block_len, block_len) ``` Here's how you can use this function: ```python s = "abbcccdddd" print(largest_block(s)) # Output: 4 ``` In this example, the largest block in the string `s` is `"dddd"`, which has a length of 4.
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输出The adjacent table of graph is:时多了->

非常抱歉,我不明白你的问题。请问你需要关于图的邻接表的输出吗?如果是的话,你可以这样输出: 假设有一个无向图,有5个顶点,边集合为:{(0,1),(0,2),(1,2),(2,3),(3,4)},邻接表可以表示为: 0: 1 2 1: 0 2 2: 0 1 3 3: 2 4 4: 3 其中每一行表示一个顶点以及与该顶点相邻的顶点列表。

What is the purpose of the “rowspan” attribute of the “td” element?

The "rowspan" attribute of the "td" element in HTML is used to specify the number of rows that a cell should span. This attribute is useful when you want to merge two or more adjacent cells in a table into a single cell that spans multiple rows. For example, if you have a table with three rows and three columns, and you want to merge the cells in the first column of the first two rows into a single cell, you can use the "rowspan" attribute. You would set the "rowspan" attribute of the first cell to "2", indicating that it should span two rows. This would cause the first two rows of the table to appear as if they have only two cells in the first column, instead of three. Here is an example of how to use the "rowspan" attribute: ``` <table> <tr> <td rowspan="2">This cell spans two rows</td> <td>Cell 1</td> <td>Cell 2</td> </tr> <tr> <td>Cell 3</td> <td>Cell 4</td> </tr> <tr> <td>Cell 5</td> <td>Cell 6</td> <td>Cell 7</td> </tr> </table> ``` In this example, the first cell spans two rows, so the first two rows of the table appear to have only two cells in the first column.

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The starting configuration of this puzzle is a row of cells, with disks located on cells through . The goal is to move the disks to the end of the row using a constrained set of actions. At each step, a disk can only be moved to an adjacent empty cell, or to an empty cell two spaces away if another disk is located on the intervening square. Given these restrictions, it can be seen that in many cases, no movements will be possible for the majority of the disks. For example, from the starting position, the only two options are to move the last disk from cell to cell , or to move the second-to-last disk from cell to cell . 1. [15 points] Write a function solve_identical_disks(length, n) that returns an optimal solution to the above problem as a list of moves, where length is the number of cells in the row and n is the number of disks. Each move in the solution should be a twoelement tuple of the form (from, to) indicating a disk movement from the cell from to the cell to. As suggested by its name, this function should treat all disks as being identical. Your solver for this problem should be implemented using a breadth-first graph search. The exact solution produced is not important, as long as it is of minimal length. Unlike in the previous two sections, no requirement is made with regards to the manner in which puzzle configurations are represented. Before you begin, think carefully about which data structures might be best suited for the problem, as this choice may affect the efficiency of your search

Create a function pixel_flip(lst, orig_lst, budget, results, i=0) that uses recursion to generate all possible new unique images from the input orig_lst, following these rules: • The input lst is the current list being processed. Initially, this will be the same as orig_lst which is the original flattened image. • The input budget represents the number of pixels that can still be flipped. When the budget reaches 0, no more pixels can be flipped. • The input results is a list of resulting flattened images with flipped pixels. Initially, this will be an empty list. • The input i represents the index of the pixel being processed, by default set to 0, which is used to drive the recursive function towards its base case (i.e., initially starting from i=0). At termination of the function, the argument results should contain all possibilities of the input orig_lst by only flipping pixels from 0 to 1 under both the budget and the adjacency constraints. fill code at #TODO def pixel_flip(lst: list[int], orig_lst: list[int], budget: int, results: list, i: int = 0) -> None: """ Uses recursion to generate all possibilities of flipped arrays where a pixel was a 0 and there was an adjacent pixel with the value of 1. :param lst: 1D list of integers representing a flattened image . :param orig_lst: 1D list of integers representing the original flattened image. :param budget: Integer representing the number of pixels that can be flipped . :param results: List of 1D lists of integers representing all possibilities of flipped arrays, initially empty. :param i: Integer representing the index of the pixel in question. :return: None. """ #TODO

Create a function pixel_flip(lst, orig_lst, budget, results, i=0) that uses recursion to generate all possible new unique images from the input orig_lst, following these rules: • The input lst is the current list being processed. Initially, this will be the same as orig_lst which is the original flattened image. • The input budget represents the number of pixels that can still be flipped. When the budget reaches 0, no more pixels can be flipped. • The input results is a list of resulting flattened images with flipped pixels. Initially, this will be an empty list. • The input i represents the index of the pixel being processed, by default set to 0, which is used to drive the recursive function towards its base case (i.e., initially starting from i=0). At termination of the function, the argument results should contain all possibilities of the input orig_lst by only flipping pixels from 0 to 1 under both the budget and the adjacency constraints. fill code at #TODO def pixel_flip(lst: list[int], orig_lst: list[int], budget: int, results: list, i: int = 0) -> None: """ Uses recursion to generate all possibilities of flipped arrays where a pixel was a 0 and there was an adjacent pixel with the value of 1. :param lst: 1D list of integers representing a flattened image . :param orig_lst: 1D list of integers representing the original flattened image. :param budget: Integer representing the number of pixels that can be flipped . :param results: List of 1D lists of integers representing all possibilities of flipped arrays, initially empty. :param i: Integer representing the index of the pixel in question. :return: None. """ #TODO def check_adjacent_for_one(flat_image: list[int], flat_pixel: int) -> bool: """ Checks if a pixel has an adjacent pixel with the value of 1. :param flat_image: 1D list of integers representing a flattened image . :param flat_pixel: Integer representing the index of the pixel in question. :return: Boolean. """ #TODO

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