Python:
# -*- coding:utf-8 -*-
class Solution:def IsPopOrder(self, pushV, popV):# write code hereq#判断需要弹出的情况的条件--压入栈的顶部和弹出栈的顶部数据相等if pushV == [] or len(pushV) != len(popV):return Nonestack = []index = 0for item in pushV:stack.append(item)while stack and stack[-1] == popV[index]:stack.pop()index += 1if stack == []:return Trueelse:return False

Python:
# -*- coding:utf-8 -*-
class Solution:def __init__(self):self.stack = []self.minvalue = []def push(self, node):# write code hereself.stack.append(node)if self.minvalue:if self.minvalue[-1] > node:self.minvalue.append(node)else:self.minvalue.append(self.minvalue[-1])else:self.minvalue.append(node)def pop(self):# write code hereif self.stack == []:return Noneself.minvalue.pop()return self.stack.pop()def top(self):# write code hereif self.stack == []:return Nonereturn self.stack[-1]def min(self):# write code hereif self.minvalue == []:return Nonereturn self.minvalue[-1]

Python
# -*- coding:utf-8 -*-
##队列是一种特殊的线性表，特殊之处在于它只允许在表的前端（front）进行删除操作，而在表的后端（rear）进行插入操作，和栈一样，队列是一种操作受限制的线性表。进行插入操作的端称为队尾，进行删除操作的端称为队头
#栈（stack）又名堆栈，它是一种运算受限的线性表。其限制是仅允许在表的一端进行插入和删除运算。这一端被称为栈顶，相对地，把另一端称为栈底。向一个栈插入新元素又称作进栈、入栈或压栈，它是把新元素放到栈顶元素的上面，使之成为新的栈顶元素；从一个栈删除元素又称作出栈或退栈，它是把栈顶元素删除掉，使其相邻的元素成为新的栈顶元素class Solution:def init(self):#添加数据栈self.acceptStack=[]#删除数据栈self.outputStack = []def push(self, node):#向添加数据的栈中添加数据self.acceptStack.append(node)def pop(self):#判断删除数据的栈中是否有数据，没有的话，就添加数据，添加数据时，要添加栈1 中删除的数据if not self.outputStack:while self.acceptStack:self.outputStack.append(self.acceptStack.pop())#如果有数据的话，就返回if self.outputStack:return self.outputStack.pop()#如果没有数据，说明没有数据添加进去，也就不需要删除数据，所以返回noneelse:return None

Python:
# -*- coding:utf-8 -*-
class Solution:# 返回对应chardef __init__(self):self.s = ""def FirstAppearingOnce(self):# write code here #method1       res=list(filter(lambda c:self.s.count(c)==1,self.s))return res[0] if res else "#"   '''
#method2  python3ret = []res = {x: self.s.count(x) for x in self.s}for key,  value in res.items():     if value == 1:ret.append(key)return ret[0] if ret else "#" #method3
for a in self.s:if self.s.count(a) == 1:return abreakreturn '#''''def Insert(self, char):# write code hereself.s  +=  char

Python:
# -*- coding:utf-8 -*-
class Solution:# s字符串def isNumeric(self, s):# write code hereif len(s) <= 0:return False# 分别标记是否出现过正负号、小数点、e，因为这几个需要特殊考虑has_sign = Falsehas_point = Falsehas_e = Falsefor i in range(len(s)):# 对于e的情况if s[i] == 'E' or s[i] == 'e':# 不同出现两个eif has_e:return False# e不能出现在最后面，因为e后面要接数字else:has_e = Trueif i == len(s) -1:return False   # 对于符号位的情况elif s[i] == '+' or s[i] == '-':# 如果前面已经出现过了符号位，那么这个符号位，必须是跟在e后面的if has_sign:if s[i-1] != 'e' and s[i-1] != 'E':return False# 如果这是第一次出现符号位，而且出现的位置不是字符串第一个位置，那么就只能出现在e后面else:has_sign = Trueif i > 0 and s[i-1] != 'e' and s[i-1] != 'E':return False# 对于小数点的情况elif s[i] == '.':# 小数点不能出现两次；而且如果已经出现过e了，那么就不能再出现小数点，因为e后面只能是整数if has_point or has_e:return False# 如果是第一次出现小数点，如果前面出现过e，那么还是不能出现小数点else:has_point = Trueif i > 0 and (s[i-1] == 'e' or s[i-1] == 'E'):return Falseelse:# 其他字符必须是‘0’到‘9’之间的if s[i] < '0' or s[i] > '9':return Falsereturn True'''##method2:try:ss=float(s)return Trueexcept:return False'''

Python:
# -*- coding:utf-8 -*-
class Solution:# s, pattern都是字符串def match(self, s, pattern):# write code hereif len(s) == 0 and len(pattern) == 0:return Trueelif len(s) != 0 and len(pattern) == 0:return Falseelif len(s) == 0 and len(pattern) != 0:# pattern中的第二个字符为*，则pattern后移两位继续比较if len(pattern) > 1 and pattern[1] == '*':return self.match(s, pattern[2:])else:return False#s与pattern都不为空的情况else:# pattern的第二个字符为*的情况if len(pattern) > 1 and pattern[1] == '*':# s与pattern的第一个元素不同，则s不变，pattern后移两位，相当于pattern前两位当成空if s[0] != pattern[0] and pattern[0] != '.':return self.match(s, pattern[2:])else:#如果s[0]与pattern[0]相同，且pattern[1]为*，这个时候有三种情况#pattern后移2个，s不变；相当于把pattern前两位当成空，匹配后面的# pattern后移2个，s后移1个；相当于pattern前两位与s[0]匹配# pattern不变，s后移1个；相当于pattern前两位，与s中的多位进行匹配，因为*可以匹配多位return self.match(s, pattern[2:]) or self.match(s[1:], pattern[2:]) or self.match(s[1:], pattern)# pattern第二个字符不为*的情况else:if s[0] == pattern[0] or pattern[0] == '.':return self.match(s[1:], pattern[1:])else:return False
C++
class Solution {
public:
/*  注意,'.'是任意一个非空字符解这题需要把题意仔细研究清楚，反正我试了好多次才明白的。首先，考虑特殊情况：1>两个字符串都为空，返回true2>当第一个字符串不空，而第二个字符串空了，返回false（因为这样，就无法匹配成功了,而如果第一个字符串空了，第二个字符串非空，还是可能匹配成功的，比如第二个字符串是“a*a*a*a*”,由于‘*’之前的元素可以出现0次，所以有可能匹配成功）,之后就开始匹配第一个字符，这里有两种可能：匹配成功或匹配失败。但考虑到pattern,下一个字符可能是‘*’， 这里我们分两种情况讨论：pattern下一个字符为‘*’或不为‘*’：1>pattern下一个字符不为‘*’：这种情况比较简单，直接匹配当前字符。如果匹配成功，继续匹配下一个；如果匹配失败，直接返回false。注意这里的“匹配成功”，除了两个字符相同的情况外，还有一种情况，就是pattern的当前字符为‘.’,同时str的当前字符不为‘0’。2>pattern下一个字符为‘*’时，稍微复杂一些，因为‘*’可以代表0个或多个。这里把这些情况都考虑到：a>当‘*’匹配0个字符时，str当前字符不变，pattern当前字符后移两位，跳过这个‘*’符号；b>当‘*’匹配1个或多个时，str当前字符移向下一个，pattern当前字符不变。（这里匹配1个或多个可以看成一种情况，因为：当匹配一个时，
由于str移到了下一个字符，而pattern字符不变，就回到了上边的情况a；当匹配多于一个字符时，相当于从str的下一个字符继续开始匹配）之后再写代码就很简单了。
*/bool match(char* str, char* pattern){if (*str == '0' && *pattern == '0')return true;if (*str != '0' && *pattern == '0')return false;//if the next character in pattern is not '*'if (*(pattern+1) != '*'){if (*str == *pattern || (*str != '0' && *pattern == '.'))return match(str+1, pattern+1);elsereturn false;}//if the next character is '*'else{if (*str == *pattern || (*str != '0' && *pattern == '.'))return match(str, pattern+2) || match(str+1, pattern);elsereturn match(str, pattern+2);}}
};

Python:
# -*- coding:utf-8 -*-
class Solution:def StrToInt(self, s):# write code herenumlist=['0','1','2','3','4','5','6','7','8','9','+','-']sum=0label=1#正负数标记if s=='':return 0for string in s:if string in numlist:#如果是合法字符if string=='+':label=1continueif string=='-':label=-1continueelse:sum=sum*10+numlist.index(string)if string not in numlist:#非合法字符sum=0break#跳出循环return sum*label

Python:
# -*- coding:utf-8 -*-
class Solution:def Add(self, num1, num2):# write code herexornum = num1 ^ num2andnum = (num1 & num2) << 1while andnum != 0:tmp1 = xornum ^ andnumtmp2 = (xornum & andnum) << 1tmp1 = tmp1 & 0xFFFFFFFF #越界检查 32位xornum = tmp1andnum = tmp2#正数返回，负数的话还原~n = -(n+1)return xornum if xornum <= 0x7FFFFFFF else ~(xornum^0xFFFFFFFF)
C++
class Solution {
public:int Add(int num1, int num2){int xornum = num1 ^ num2;int andnum = (num1 & num2) << 1;while(andnum != 0){int tmp1 = xornum ^ andnum;int tmp2 = (xornum & andnum) << 1;xornum = tmp1;andnum = tmp2;}return xornum;}
};

Python:
# -*- coding:utf-8 -*-
class Solution:def Sum_Solution(self, n):# write code here#return sum(list(range(1,n+1)))return (pow(n,2) + n) >> 1

Python:
# -*- coding:utf-8 -*-
class Solution:def IsContinuous(self, numbers):# write code here#method1if len(numbers):while min(numbers) == 0:numbers.remove(0)if max(numbers) - min(numbers) <= 4 and len(numbers) == len(set(numbers)):return Truereturn False
C++
class Solution {
public:bool IsContinuous( vector<int> numbers ) {if(numbers.size()!=5) return false;sort(numbers.begin(),numbers.end());int i=0;while(numbers[i]==0) i++;if(numbers[4]-numbers[i]>4) return false;for(int j=i;j<4;j++){if(numbers[j]==numbers[j+1]) return false;}return true;}
};

Python:
# -*- coding:utf-8 -*-
class Solution:def ReverseSentence(self, s):# write code hereif len(s) == 0:return ss = s.split(" ")#print(s[::-1])res = " ".join(s[::-1])return res

Python:
# -*- coding:utf-8 -*-
class Solution:def LeftRotateString(self, s, n):# write code here     l = len(s)if l <= 0 or n < 0:return sn = n % la = s[:n][::-1]b = s[n:][::-1]c =  b[::-1] + a[::-1]return c'''return s[n:] + s[:n]'''

Python
# -*- coding:utf-8 -*-
class Solution:def FirstNotRepeatingChar(self, s):# write code here#return [i for i in range(len(s)) if s.count(s[i])==1][0] if s else -1'''if s:a = [i for i in range(len(s)) if s.count(s[i])==1][0]return a else:return -1'''#判断输入条件if len(s)<=0 or len(s)>10000:return -1#count用于统计字符串中某个字符的出现]数#index为计算字符串中某个字符的位置for i in s:if s.count(i)==1:return s.index(i)break

Python
# -*- coding:utf-8 -*-
class Solution:def Power(self, base, exponent):# write code here#return base**exponent   if base == 0:return 0if exponent == 0:return 1p = abs(exponent)res = 1while(p > 0):#如果最后一位为1，那么给res乘上这一位的结果if (p & 1 == 1):res =res * basep = p >> 1base = base * basereturn res if exponent > 0 else 1/res
C++
class Solution {
public:double Power(double base, int exponent) { //codeif(base == 0.0)return 0;if(exponent == 0.0)return 1;long long p = abs((long long)exponent);double r = 1.0;while(p){if(p & 1)  r *= base;base *= base;p >>= 1;}return exponent < 0 ? 1/ r : r;}
};

Python:def NumberOf1(self, n):# write code here#flag = 1count = 0if n < 0: ##必须加n = n & 0xffffffff #为了获得负数（十进制表示）的补码，需要手动将其和十六进制数0xfffffffd进行按位与操作while n:count += 1n = (n-1)&nreturn count
C++
class Solution {
public:int  NumberOf1(int n) {/*int count = 0;int flag = 1;//C语言中 << 是逻辑移位，不是循环移位; 1左移32位后为0while(flag){if(n&flag)count++;flag = flag << 1;}return count;*/int count = 0;while(n){count ++;n = (n-1) & n;//消除n最右边的一个0}return count;}
};

Python
# -*- coding:utf-8 -*-
class Solution:# s 源字符串def replaceSpace(self, s):# write code here#s1 = s.replace(' ', '%20')#return s1new_s = ''for j in s:if j == ' ':new_s = new_s + '%20'else:new_s = new_s + jreturn new_s
C++
class Solution {
public:void replaceSpace(char *str,int length) {if(str==NULL||length<0)return;int i=0;int oldnumber=0;//记录以前的长度int replacenumber=0;//记录空格的数量while(str[i]!='0'){oldnumber++;if(str[i]==' '){replacenumber++;}i++; }int newlength=oldnumber+replacenumber*2;//插入后的长度if(newlength>length)//如果计算后的长度大于总长度就无法插入return;int pOldlength=oldnumber; //注意不要减一因为隐藏个‘0’也要算里int pNewlength=newlength;while(pOldlength>=0&&pNewlength>pOldlength)//放字符{if(str[pOldlength]==' ') //碰到空格就替换{str[pNewlength--]='0';str[pNewlength--]='2';str[pNewlength--]='%';}else //不是空格就把pOldlength指向的字符装入pNewlength指向的位置{str[pNewlength--]=str[pOldlength];}pOldlength--; //不管是if还是elsr都要把pOldlength前移}}
};

Python
# -*- coding:utf-8 -*-
class Solution:def NumberOf1Between1AndN_Solution(self, n):# write code here#return ''.join(map(str, range(n+1))).count('1')#method 1         # 将1-n全部转换为字符串# 只需要统计每个字符串中'1'出现的次数并相加即可     return ''.join(map(str,range(n+1))).count('1')
C++
class Solution {
public:int NumberOf1Between1AndN_Solution(int n)
{//方法1：暴力求解//int count = 0;// for(int i=0; i<=n; i++){// int temp = i;// while(temp){//  if(temp%10 == 1){//    count++;// }// temp /= 10;//  }//}//return count; //方法2int count = 0;long long m = 1;for(m=1; m<=n; m*=10){int a = n/m;int b = n%m;count += (a+8)/10*m + (a%10==1)*(b+1);}return count;}
};
//举一反三：如果求2的个数
//count += (a+7)/10*m + (a%10==2)*(b+1); 以此类推