while ~isempty(stack) node = stack(end); stack(end) = []; if visited(node) == 0 visited(node) = 1; timestamps(node) = timestamp; timestamp = timestamp + 1; % 这里可以根据实际需求进行相应的操作,例如输出节点值 disp(['Visited Node ', num2str(node)]); neighbors = find(graph(node, :)); % 获取相邻节点 for i = length(neighbors):-1:1 if visited(neighbors(i)) == 0 stack(end+1) = neighbors(i); end end end end disp('Timestamps:'); disp(timestamps); 结束里边有错误visited = Zeros(1, length(graph)); 这个问题
时间: 2024-04-08 19:34:32 浏览: 34
非常抱歉,我之前给出的回答错误。问题出现在第2行的`Zeros`函数调用中,应该是`zeros`(小写字母)而不是`Zeros`(大写字母)。在MATLAB中,函数名是区分大小写的,因此应该使用小写字母来调用`zeros`函数。
因此,将第2行的代码修改为以下内容应该可以解决问题:
```matlab
visited = zeros(1, length(graph));
```
请注意,`zeros`函数用于创建一个全零的向量或矩阵,与您在深度优先搜索算法中的需求相符。非常抱歉给您带来困扰,请尝试修改代码并再次运行。
相关问题
function dfsWithTimestamp(graph, startNode) visited = zeros(1, length(graph)); visited = zeros(1, length(graph)); timestamp = 1; stack = startNode; timestamps = zeros(1, length(graph)); while ~isempty(stack) node = stack(end); stack(end) = []; if visited(node) == 0 visited(node) = 1; timestamps(node) = timestamp; timestamp = timestamp + 1; % 这里可以根据实际需求进行相应的操作,例如输出节点值 disp(['Visited Node ', num2str(node)]); neighbors = find(graph(node, :)); % 获取相邻节点 for i = length(neighbors):-1:1 if visited(neighbors(i)) == 0 stack(end+1) = neighbors(i); end end end end disp('Timestamps:'); disp(timestamps); end 结束里边有错误visited = zeros(1, length(graph)); 这个问题
非常抱歉,我之前的回答中有误导。我再次检查了您提供的代码,发现确实存在问题。错误出现在第3行和第4行,这两行都是将`visited`数组初始化为全零向量。
您应该只保留其中一行初始化代码,并删除另一行。以下是已经修复的代码示例:
```matlab
function dfsWithTimestamp(graph, startNode)
visited = zeros(1, length(graph));
timestamp = 1;
stack = startNode;
timestamps = zeros(1, length(graph));
while ~isempty(stack)
node = stack(end);
stack(end) = [];
if visited(node) == 0
visited(node) = 1;
timestamps(node) = timestamp;
timestamp = timestamp + 1;
% 这里可以根据实际需求进行相应的操作,例如输出节点值
disp(['Visited Node ', num2str(node)]);
neighbors = find(graph(node, :)); % 获取相邻节点
for i = length(neighbors):-1:1
if visited(neighbors(i)) == 0
stack(end+1) = neighbors(i);
end
end
end
end
disp('Timestamps:');
disp(timestamps);
end
```
请注意,我已经删除了第3行中的重复初始化代码。非常抱歉给您带来困惑,并感谢您的指正。请尝试使用修复后的代码并再次运行。
class Solution { public String removeDuplicateLetters(String s) { int[] tem = new int[26]; for (int i = 0; i < s.length(); i++) { char tems = s.charAt(i); tem[tems - 'a']++; } boolean[] visited = new boolean[26]; Stack<Character> stack = new Stack<>(); for (int i = 0; i < s.length(); i++) { char get = s.charAt(i); while (!stack.isEmpty() && stack.peek() > get && !visited[get - 'a'] && tem[stack.peek() - 'a'] > 0) { visited[stack.peek() - 'a'] = false; stack.pop(); } if (visited[get - 'a']) { tem[get - 'a']--; continue; } visited[get - 'a'] = true; stack.push(get); tem[get - 'a']--; } StringBuilder sBuilder = new StringBuilder(); while (!stack.isEmpty()) { sBuilder.append(stack.pop()); } return sBuilder.reverse().toString(); } }帮我写完整的代码
这是一个去除字符串中重复字母的算法,下面是完整的代码实现:
```
public class Solution {
public String removeDuplicateLetters(String s) {
int[] tem = new int[26];
for (int i = 0; i < s.length(); i++) {
char tems = s.charAt(i);
tem[tems - 'a']++;
}
boolean[] visited = new boolean[26];
Stack<Character> stack = new Stack<>();
for (int i = 0; i < s.length(); i++) {
char get = s.charAt(i);
while (!stack.isEmpty() && stack.peek() > get && !visited[get - 'a'] && tem[stack.peek() - 'a'] > 0) {
visited[stack.peek() - 'a'] = false;
stack.pop();
}
if (visited[get - 'a']) {
tem[get - 'a']--;
continue;
}
visited[get - 'a'] = true;
stack.push(get);
tem[get - 'a']--;
}
StringBuilder sBuilder = new StringBuilder();
while (!stack.isEmpty()) {
sBuilder.append(stack.pop());
}
return sBuilder.reverse().toString();
}
}
```
该算法的具体思路是用一个栈来维护一个字典序最小的字符串,并且保证每个字符只出现一次。对于每个字符,先判断它是否已经在栈中出现过,如果出现过,那么不需要考虑,直接跳过;否则,需要判断栈顶元素和当前元素的大小关系,如果栈顶元素大于当前元素且栈顶元素在后面还会出现,那么就需要将栈顶元素弹出,然后再判断新的栈顶元素和当前元素的大小关系,直到栈顶元素小于等于当前元素为止。最后将栈中元素弹出并倒序输出即可。
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