leetcode习题练习

day001 两数之和

#!user/bin/env python
# -*- coding:utf-8 -*-
#方法1
def Sum(nbs,tgt):
    len_nums = len(nbs)
    for i in range(len_nums):
        for j in range(i+1,len_nums):
            if nbs[i] + nbs[j] == tgt:
                return [i,j]
print(Sum([2,7,11,15],9))

#方法2 二遍哈希表
def Sum(nums,tgt):
    len_num = len(nums)
    for i in range(len_num):
        dif = tgt - nums[i]
        if dif in nums[i+1:]:
            return [i,nums[i+1:].index(dif)+i+1]
print(Sum([2,7,11,15],9))

#方法3 一遍哈希表
def Sum(nums,tgt):
    len_num = len(nums)
    for i in range(len_num):
        dif = tgt - nums[i]
        if dif in nums[:i]:
            return [nums[:i].index(dif),i]
print(Sum([2,7,11,15],9))

day002 给定一个字符串，找出其中不含有重复字符的最长子串的长度

# global num
# 求出字符串最长长度，一次，无迭代(代码正确)，求出最大值，代码错误！而且题目理解错误，只要求出数字即可！！！
# def func_str(x_li):
#     num = 0
#     for i in range(len(x_li)):
#         li1.append(x_li[i])
#         if num == 0:
#             num = num + 1
#             continue
#         for j in range(len(li1)-1):
#             if x_li[i] == li1[j]:
#                 return li1[:-1]  #返回值形如['a','b','c']
# global num
# li1 = []
# li = list(input('>>>'))
# str_max = func_str(li)
# print(''.join(str_max))

#以下错误代码
def func_str(x_li):
    num = 0
    for i in range(len(x_li)):
        li1.append(x_li[i])
        if num == 0:
            num = num + 1
            continue
        for j in range(len(li1)-1):
            if x_li[i] == li1[j]:
                return li1[:-1]  #返回值形如['a','b','c']
global num
li1 = []
li = list(input('>>>'))
str_max = func_str(li)
print(''.join(str_max)) # 输入aaa 输出a
s = len(str_max)
while True:
    li1.clear()
    str_max1 = func_str(list(set(li)-set(str_max)))
    # print(li2)
    # li2 = func_str(li[s:-1])
    if len(str_max1) > s:
        str_max = str_max1
    print(''.join(str_max))

#reference code 14行精简代码！！
def func_str(s):
    max_len = 0
    if s is None or len(s) == 0:
        return max_len
    str_dict = {}
    one_max = 0
    start = 0
    for i in range(len(s)):
        if (s[i] in str_dict) and (str_dict[s[i]] >= start): #字典中的数要从start算起，并且s[i]存在于i以后的字典中
            start = i
            print(str_dict[s[i]])  #测试用
        one_max = i - start + 1
        str_dict[s[i]] = i  #字典不允许key，value重复
        # print(str_dict[s[i]])
        # print(str_dict[s[i]])
        max_len = max(max_len, one_max)
    return max_len
print(func_str('abc123abcefghi1231234567890zxcvbnmasdfghjkl'))

day003 整数反转，给出一个32位的有符号整数，你需要将这个整数中每位上的数字进行反转

str = input('>>>')
li = []
for i in str:
    li.append(i)
if li[0] is '-':
    li.pop(0)
    li.reverse()
    print(''.join(li),end='')
    print('-')
else:
    li.reverse()
    print(li)
    for i in li:
        if i == '':
            li.pop(int(i))
    # print(li)
        elif int(i) is not 0:
            print('yes')
            print(''.join(li))
            break

day004 最长回文字串：给定一个字符串s，找到s中最长的回文子串。假设s的最大长度为1000. 回文子串定义：如level,noon

#判断一个固定长度字符串是否回文的3中方法
方法1_字符串逆序法
def is_palindrome1(text):
    l = list(text)
    l.reverse()
    t1 = ''.join(l)
    if t1 == text:
        print 'the text is palindrome'
    else:
        print 'the text is not palindrome'

方法2_字符串逆序索引
def is_palindrome2(text):
    t1 = text[::-1]
    if t1 == text:
        print 'the text is palindrome'
    else:
        print 'the text is not palindrome'

方法3_传统法，s[len(s)-x-1] == s[x] ??
def is_palindrome3(text):
    r = True
    for x in range(len(text)):
        print x,text[x]
        if text[x] != text[len(text)-x-1]:
            print 1
            r = False
            break
    if r == True:
        print 'the text is palindrome'
    else:
        print 'the text is not palindrome'

#中心扩展法 代码好像有点问题，3.5revision!
def huiwen(str):
    str_max = ''
    for i in range(2*len(str)-1):
        #偶数情况 如在12321中i==2，为偶数情况
        if i % 2 == 0:
            start = end = i // 2 #/整除
            while start >= 0 and end < len(str) and str[start] == str[end]:
                start -= 1
                end += 1
        #奇数情况
        else:
            start = (i-1)//2
            end = (i-1)//2
            while start >= 0 and end <= len(str) and str[start] == str[end]:
                start -= 1
                end += 1
        if len(str) <= (end-start+1):
            str_max = str[start+1:end]
    return str_max
print(huiwen('cbbd'))

本代码编写分两步，1向字符串中插入'#',字符串首尾+‘#’；2字符串中以每个元素为中心进行扩展，查找最大回文字符串。

def max_huiwen(string):
    li = [i for i in string]
    newli_list = '#' + '#'.join(li) + '#'
    max_str = 0
    length = len(newli_list)
    for index in range(0,length):
        max_substr = get_length(newli_list,index)
        if max_substr > max_str:
            max_str = max_substr
    return max_str
def get_length(s_li,index):
    length = len(s_li)
    submax_str = 0
    for i in range(1,index+1):
        if index+i < length and s_li[index-i] == s_li[index+i]:
            submax_str = submax_str + 1
        else:
            break
    return submax_str
res = max_huiwen('13553143211234321123aaaaaaaaaaaaaaaaaa')
print(res)
day005 有两个有序数组，求他们的中位数，如num1=[1,3]，num2=[2,4] 中位数是（2+3）/2=2.5；再如num1=[1,3],num2=[2] 中位数是2.

num1 = [1,3,5,6,7,9,11,22]
num2 = [2,4,5]
def medium1(num1):
    # num = num1 + num2
    # sorted(num)
    length1 = len(num1)

    if length1 % 2 == 0: #length1长度为偶数
        res1 = (num1[length1//2] + num1[length1//2-1])/2
        return res1
    else: #length1为奇数
        res1 = num1[(length1-1)//2]
        return res1
def medium2(num1):
    length2 = len(num2)
    if length2 % 2 == 0: #length2长度为偶数
        res2 = (num2[length2//2] + num2[length2//2-1])/2
        return res2
    else: #length2为奇数
        res2 = num2[(length2-1)//2]
        return res2
a = medium1(num1)
b = medium1(num2)
print((a+b)/2)

day005 将罗马字符转换为阿拉伯数字：算法，字符串左边小于右边，累加；左边大于右边，左边数字 - 2*右边数字

def func(roman_substr):
    res = 0
    dic = {
        'I':1,
        'V':5,
        'X':10,
        'L':50,
        'C':100,
        'D':500,
        'M':1000
    }
    for i in range(0,len(roman_substr)):
        if i == 0 or dic[roman_substr[i]] <= dic[roman_substr[i-1]]:
            res += dic[roman_substr[i]]
        else:
            res += dic[roman_substr[i]] - 2*dic[roman_substr[i-1]]
    return res
roman_str = ['DII','XV','MDCLXVI','XII','VIII','XCIX','XII']
for roman_substr in roman_str:
    print(func(roman_substr))
day007 输入的一个字符串和总行数，按照zigzag类型输出，如输入‘HKTSUOFFER’,3；输出HUEKTOFRSF(18行代码！)

def zigzag(s, nRows):
    if s == '' or nRows <= 1:
        return s
    li = []
    size = 2*nRows-2
    for i in range(nRows):
        j = i #values of lines from this one
        if i == 0 or i == nRows-1:
            while j < len(s):
                li.append(s[j])
                j +=size #algorithm1 for initial and end lines
        else:
            while j < len(s):
                li.append(s[j])
                j += size-2*i #algorithm2 for uninitial and unend lines
    return li
res = zigzag('HKSTUOFFER', 3)
print((''.join(res)).upper())

day006 Regular Expression Matching, 规则1 '.'可以匹配任何字母，‘*’可以重复一次‘*’之前的一个字母；规则2 输入两个字符串a和b，如b能覆盖a，返回True，否则返回False. 如1输入’ab‘和‘a*’ 返回False. 如2输入'aa'和‘a*’ 返回True.

def judge_twostr():
    str = input('输入原字符串>>>')
    str_li = []
    str_str1 = ''
    s = input('输入想要匹配的字符串>>>')
    s_li = []
    str_str2 = ''
    for i in range(len(str)):
        str_li.append(str[i])
    for j in range(len(s)):
        s_li.append(s[j])
    for k in range(len(s_li)):
        if s_li[k] == '.': #字符串中部遇到字符'.'
            s_li[k] = str_li[k]
        if s_li[k] == '*': #字符串中部遇到字符‘*’
            s_li[k] = s_li[k-1]
    str_str1 += ''.join(str_li) #str_li和s_li转换为字符串
    str_str2 += ''.join(s_li)  #str_li 范围小    s_li范围大
    if str_str2 in str_str1 or str_str1 == str_str2:
        print('T')
    else:
        print('F')
if __name__ == '__main__':
    judge_twostr()
# 复习字符串处理
def judge_twostr():
    str = input('输入原字符串>>>')
    li_str = list(str)
    str_str1 = ''
    s = input('输入想要匹配的字符串>>>')
    li_s = list(s)
    str_str2 = ''
    for k in range(len(li_s)):
        if li_s[k] == '.': #字符串中部遇到字符'.'
            li_s[k] = li_str[k]
        if li_s[k] == '*': #字符串中部遇到字符‘*’
            li_s[k] = li_s[k-1]
    str_str1 += ''.join(li_str) #str_li和s_li转换为字符串
    str_str2 += ''.join(li_s)  #str_li 范围小    s_li范围大
    if str_str2 in str_str1 or str_str1 == str_str2:
        print('T')
    else:
        print('F')
if __name__ == '__main__':
    judge_twostr()