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.

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:

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:

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.

相关问题

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

Here's a possible implementation of the `solve_identical_disks` function using breadth-first graph search:

from collections import deque

def solve_identical_disks(length, n):
    # Initialize the starting configuration
    start = [True] * n + [False] * (length - n)

    # Define a function to generate all possible successor configurations
    def successors(config):
        for i in range(length):
            if not config[i]:
                if i + 1 < length and config[i + 1]:
                    # Move disk to adjacent empty cell
                    yield config[:i] + [True] + [False] + config[i + 2:]
                elif i + 2 < length and config[i + 2]:
                    # Move disk to empty cell two spaces away
                    yield config[:i] + [True] + config[i + 2:i + 3] + [False] + config[i + 3:]

    # Perform breadth-first graph search to find the goal configuration
    queue = deque([(start, [])])
    visited = set([tuple(start)])
    while queue:
        config, moves = queue.popleft()
        if config == [False] * (length - n) + [True] * n:
            return moves
        for successor in successors(config):
            if tuple(successor) not in visited:
                queue.append((successor, moves + [(config.index(True), successor.index(True))]))
                visited.add(tuple(successor))

    # If the goal configuration is not reachable, return None
    return None

The `start` variable represents the starting configuration as a list of `True` values for the cells occupied by disks and `False` values for the empty cells. The `successors` function generates all possible successor configurations for a given configuration by moving a disk to an adjacent empty cell or to an empty cell two spaces away. The `queue` variable is used to store the configurations to be explored, along with the list of moves required to reach them. The `visited` set is used to keep track of the configurations that have already been explored, in order to avoid revisiting them.

The function returns the list of moves required to reach the goal configuration, which is represented as a list of `False` values for the cells before the disks and `True` values for the cells occupied by the disks. If the goal configuration is not reachable, the function returns `None`.

Vika and her friends went shopping in a mall, which can be represented as a rectangular grid of rooms with sides of length n and m. Each room has coordinates (a, b), where 1 ≤ a ≤ n, 1 ≤ b ≤ m. Thus we call a hall with coordinates (c, d) a neighbouring for it if ｜a-c｜+|b-d｜=1.Tired of empty fashion talks, Vika decided to sneak away unnoticed. But since she hasn't had a chance to visit one of the shops yet, she doesn't want to leave the mall. After a while, her friends noticed Vika's disappearance and started looking for her.Currently, Vika is in a room with coordinates (z, y), and her k friends are in rooms with coordinates (x1, Y1), (x2, 32),..,(ük) 3k).respectively. The coordinates can coincide. Note that all the girls must move to the neighbouring rooms.Every minute, first Vika moves to one of the adjacent to the side rooms of her choice, and then each friend (seeing Vika's choice) also chooses one of the adjacent rooms to move to.If at the end of the minute (that is, after all the girls have moved on to the neighbouring rooms) at least one friend is in the same room as Vika, she is caught and all the other friends are called.Tell us, can Vika run away from her annoying friends forever, or will she have to continue listening to empty fashion talks after some time?如果能不被抓住输出yes，否则输出no，给出c++代码及中文解释

可以使用深度优先搜索（DFS）来解决这个问题。首先，我们可以将问题抽象为一个图，其中每个房间都是一个节点，相邻房间之间有边相连。

首先，我们需要定义一个辅助函数，用于检查给定的坐标是否在合法的范围内：

bool isValid(int x, int y, int n, int m) {
    return (x >= 1 && x <= n && y >= 1 && y <= m);
}

接下来，我们可以使用DFS来遍历所有可能的移动方式。我们从Vika的起始位置开始，依次向四个方向移动，然后递归地对每个朋友进行相同的操作。如果在递归过程中，任何一个朋友与Vika在同一个房间，则说明Vika会被抓住。

以下是完整的C++代码实现：

#include <iostream>
#include <vector>

using namespace std;

bool isValid(int x, int y, int n, int m) {
    return (x >= 1 && x <= n && y >= 1 && y <= m);
}

bool dfs(int vx, int vy, vector<pair<int, int>>& friends, vector<vector<bool>>& visited, int n, int m) {
    if (vx == 0 && vy == 0) {
        // Vika has successfully escaped
        return true;
    }
    
    visited[vx][vy] = true;
    
    // Possible neighbors' coordinates
    vector<pair<int, int>> neighbors = {{vx-1, vy}, {vx+1, vy}, {vx, vy-1}, {vx, vy+1}};
    
    for (auto neighbor : neighbors) {
        int nx = neighbor.first;
        int ny = neighbor.second;
        
        if (isValid(nx, ny, n, m) && !visited[nx][ny]) {
            bool caught = false;
            
            for (auto friendCoord : friends) {
                int fx = friendCoord.first;
                int fy = friendCoord.second;
                
                if (nx == fx && ny == fy) {
                    caught = true;
                    break;
                }
            }
            
            if (!caught && dfs(nx, ny, friends, visited, n, m)) {
                return true;
            }
        }
    }
    
    return false;
}

string canVikaEscape(int n, int m, int vx, int vy, vector<pair<int, int>>& friends) {
    vector<vector<bool>> visited(n+1, vector<bool>(m+1, false));
    
    if (dfs(vx, vy, friends, visited, n, m)) {
        return "Yes";
    } else {
        return "No";
    }
}

int main() {
    int n, m, vx, vy, k;
    cin >> n >> m >> vx >> vy >> k;
    
    vector<pair<int, int>> friends(k);
    
    for (int i = 0; i < k; i++) {
        cin >> friends[i].first >> friends[i].second;
    }
    
    cout << canVikaEscape(n, m, vx, vy, friends) << endl;
    
    return 0;
}

该代码首先读取输入的n，m，vx，vy和k值。接下来，它读取k个朋友的坐标，并调用canVikaEscape函数来判断Vika是否能逃脱。最后，它输出结果。

希望这可以帮助到你！