python：remove，pop和del方法的区别（python中remove和pop的区别）

25-04-27 4

本篇文章给大家谈谈python：remove，pop和del方法的区别，以及python中remove和pop的区别的知识点，同时本文还将给你拓展301.RemoveInvalidParenthese

本篇文章给大家谈谈python：remove，pop和del方法的区别，以及python中remove和pop的区别的知识点，同时本文还将给你拓展301. Remove Invalid Parentheses 去掉不合理的括号、316. Remove Duplicate Letters and 402. Remove K Digits、centos – 如何撤消“yum remove python”、centos – 服务器在“yum remove python”之后擦除等相关知识，希望对各位有所帮助，不要忘了收藏本站喔。

本文目录一览：

python：remove，pop和del方法的区别（python中remove和pop的区别）
301. Remove Invalid Parentheses 去掉不合理的括号
316. Remove Duplicate Letters and 402. Remove K Digits
centos – 如何撤消“yum remove python”
centos – 服务器在“yum remove python”之后擦除

python：remove，pop和del方法的区别（python中remove和pop的区别）

remove
remove(item)方法是直接对可变序中的元素进行检索删除，返回的是删除后的列表,不返回删除值（返回None）

list1=[1,3,6,7,8]
print list1.remove(3) #对列表元素进行搜索删除，而不是下表
print list1

None
[1,8]

pop
pop(index)方法是对可变序列中元素下标进行检索删除，返回删除值

list1=[1,8]
print list1.pop(3),#对列表下表进行检索删除
print list1

7
[1,8]

dict1={‘Abby‘:21,‘Bob‘:22,‘cindy‘:21}
print dict1.pop(1)
print dict1

del
del(list[index])方法是对可变序列中元素下边进行检索删除，不返回删除值

list1=[1,8]
del list[3]
print list1

[1.3,8]

301. Remove Invalid Parentheses 去掉不合理的括号

Remove the minimum number of invalid parentheses in order to make the input string valid. Return all possible results.

Note: The input string may contain letters other than the parentheses ( and ).

Examples:

"()())()" -> ["()()()", "(())()"] "(a)())()" -> ["(a)()()", "(a())()"] ")(" -> [""]

解题过程：

用BFS，因为这样的话，对于一个初始s,遍历它，并减去一个不合理的符号，得到的第一层就是全是n-1长度的，当减去一些后得到一个合理的字符串时，停止剪。那么这个层数就是减去的个数也是最小减去个数得到的结果。

当root层时s有长度n,则到第二层时，有c(n,n-1)个长度为n-1的子串，同时进行验证是时间为O(n-1)，因此第一层的时间复杂度为O(n-1)*c(n,n-1);同理第二层的时间复杂度为O(n-2)*c(n-1,n-2)。。。。

一共的时间复杂度为

T(n) = n x C(n, n) + (n-1) x C(n, n-1) + ... + 1 x C(n, 1) = n x 2^(n-1).

注意下面代码有个地方要注意：

就是当验证到正确的字符串时，要使found等于true,因为这样的话，就等于不用再对队列中的字符串进行剪了，只要把队列中剩下的字符串进行验证就行了。

开始时，我将continue写到if(isvalid(str))的范围里了，这样的话，当队列中的其他字符串取出进行验证发现不正确时，还会继续往下减，这样就不对了，因为已经得到了进行最少步剪的步数的到的结果了。

代码如下：

class Solution {

public:

Given a string which contains only lowercase letters, remove duplicate letters so that every letter appear once and only once. You must make sure your result is the smallest in lexicographical order among all possible results. Given "bcabc" Return "abc" Given "cbacdcbc" Return "acdb"

每个字母只留一个，而且保证字典序最小。从前往后扫描，还要往前看一个检出是否删除的，要用stack解。stack里记录的就是待用的字母。首先要有一个counter记录下每个字母的总个数，stack删除字母的时候如果后面还有余量，则可以放心删除。还有个关键词: once and only once。需要一个数据结构记录是否使用这个字母，可以用boolean。

public class Solution { public String removeDuplicateLetters(String s) { ArrayDeque<Character> stk = new ArrayDeque<Character>(); int[] counter = new int[26]; boolean[] visited = new boolean[26]; char[] chs = s.toCharArray(); for(char c: chs){ counter[c - ''a'']++; } for(char c : chs){ counter[c - ''a'']--; if(visited[c - ''a'']) continue; while(!stk.isEmpty() && stk.peekLast() > c && counter[stk.peekLast() -''a''] > 0){ visited[stk.peekLast() - ''a''] = false; stk.pollLast(); } stk.addLast(c); visited[c-''a''] = true; } StringBuilder sb = new StringBuilder(); for(char c: stk){ sb.append(c); } return sb.toString(); } }

Input: num = "1432219", k = 3 Output: "1219" Explanation: Remove the three digits 4, 3, and 2 to form the new number 1219 which is the smallest. Input: num = "10200", k = 1 Output: "200" Explanation: Remove the leading 1 and the number is 200. Note that the output must not contain leading zeroes. Input: num = "10", k = 2 Output: "0" Explanation: Remove all the digits from the number and it is left with nothing which is 0.

public class Solution { public String removeKdigits(String num, int k) { int digit = num.length() - k; char[] stk = new char[num.length()]; int top = 0; for(char c: num.toCharArray()){ while(top > 0 && stk[top-1] > c && k > 0){ k--; top--; } stk[top++] = c; } int idx = 0; while(idx < digit && stk[idx] == ''0'') idx++; return idx == digit ? "0" : new String(stk, idx, digit-idx); } }

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