leetcode刷题_Python(1-10)
一个小小白。。开始leetcode刷题。
是跟着这个视频学的。https://space.bilibili.com/405863549/video
小姐姐讲的很好。易懂。然后我跟着敲了一遍代码。记录下来。
- 第一题：两数之和（简单）
找到一个数组中两个值得和等于固定值（target）返回这俩值的索引
例子：
“”"
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
“”"
（1）第一种解法。直白。但耗时多
class Solution: def twoSum1(self, nums, target): for i in nums: j = target - i start_index = nums.index(i) next_index = start_index + 1 temp_index = nums[next_index:] if j in temp_index: return [nums.index(i),next_index+temp_index.index(j)]
(2) 第二种解法。方便。
class Solution: def twoSum(self, nums, target) : dict ={} for i in range(len(nums)): if target - nums[i] not in dict: dict[nums[i]] = i else: return [dict[target-nums[i]],i]
2.第七题：（简单）：Reverse Integer 转置整数
Given a 32-bit signed integer, reverse digits of an integer.
（1）Example 1:
Input: 123
Output: 321
（2）Example 2:
Input: -123
Output: -321
需要注意的是，转置后的数不能溢出32位整数范围：[−2^31, 2^31 − 1];也就是[-2147483648,2147483647]
代码：
(1) 不推荐
class Solution:
def reverse(self, x):
num = 0 #将输出的数设为0
a = abs(x)
while(a != 0):
temp = a % 10 #取出个位数
num = num*10 + temp
a = int(a/10)
if x > 0 and num < 2147483647:
return num
elif x < 0 and num < 2147483648:
return -num
else:
return 0
（2）采用字符串的方法转置，简单。推荐
class Solution:
def reverse(self, x):
s = str(x)[::-1].rstrip("-")
if int(s) < 2**31:
if x > 0:
return int(s)
else:
return 0 - int(s)
return 0
- 第九题（简单）：Palindrome number
Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.
确定一个数是不是回文数。回文就是顺读和倒读都一样的词句
（1）Example 1:
Input: 121
Output: true
（2）Example 2:
Input: -121
Output: false
（3）Example 3:
Input: 10
Output: false
代码：
(1) 复杂，不推荐
class Solution:
def isPalindrome(self, x):
num = 0
a = abs(x)
while(a!=0):
temp = a % 10
num = num *10 + temp
a = a // 10
if x >= 0 and x == num:
return True
else:
return False
（2）采用字符串方法。简单
class Solution:
def isPalindrome(self, x):
if x < 0:
return False
s = str(x)[::-1]
if int(s) == x:
return True
else:
return False
4.第13个题（简单）：
Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.
Symbol Value
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
一般罗马数字都是按照，从左到右，从大到小排列的。但也有比如，后面一位代表的数，比前面一位大的时候，就用后面一位减去前面一位。
比如：
I can be placed before V (5) and X (10) to make 4 and 9.
X can be placed before L (50) and C (100) to make 40 and 90.
C can be placed before D (500) and M (1000) to make 400 and 900.
例子：
（1）Example 1:
Input: “III”
Output: 3
（2）Example 2:
Input: “IV”
Output: 4
（3）Example 3:
Input: “IX”
Output: 9
（4）Example 4:
Input: “LVIII”
Output: 58
Explanation: L = 50, V= 5, III = 3.
（5）Example 5:
Input: “MCMXCIV”
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
代码：
class Solution:
def romanToInt(self, s):
result = 0
numeral_map = {"I" : 1, "V": 5, "X": 10, "L": 50, 'C': 100,\
"D": 500, "M" :1000}
for i in range(len(s)):
if i >0 and numeral_map[s[i]] > numeral_map[s[i-1]]:
result += numeral_map[s[i]] - 2 * numeral_map[s[i-1]]
else:
result += numeral_map[s[i]]
return result
以例5为例，因为后面一位减去前面一位的时候（M-C），前面一位已经加过了（变成：M+C+M-C），但我们想要得到的结果是M+(M-C），所以要再减去1次C，就是前面的数。所以代码中要×2
5 .第14题（简单）：Longest Common Prefix 最大的相同前缀
Write a function to find the longest common prefix string amongst an array of strings.
If there is no common prefix, return an empty string “”.
（1）Example 1:
Input: [“flower”,“flow”,“flight”]
Output: “fl”
（2）Example 2:
Input: [“dog”,“racecar”,“car”]
Output: “”
Explanation: There is no common prefix among the input strings.
Note:
All given inputs are in lowercase letters a-z.
代码：
有两种解法：
① 第一种解法：就是用第一个数作为参考，后面的数不断和第一个数比较。
class Solution:
def longestCommonPrefix(self, strs):
if not strs: # 如果没有strs
return ""
for i in range(len(strs[0])):
for string in strs[1:]:
if i >= len(string) or strs[0][i] != string[i]:
return strs[0][:i]
return strs[0] #有strs，但里面的符号是空
② 第二种解法：用到 zip 和 set
class Solution(object):
def longestCommonPrefix(self, strs):
"""
:type strs: List[str]
:rtype: str
"""
result = ""
zzip = zip(*strs)
# 解压，返回一个由各个元祖组成的数组。长度为strs中最小序列的长度
for temp in zzip:
temp_set = set(temp)
if len(temp_set) == 1:
result += temp[0]
else:
break
return result
- 第20题 （简单）：Valid Parentheses 合法的括号
Given a string containing just the characters ‘(’, ‘)’, ‘{’, ‘}’, ‘[’ and ‘]’,
determine if the input string is valid.
An input string is valid if:
Open brackets must be closed by the same type of brackets.
Open brackets must be closed in the correct order. 正确的顺序
Note that an empty string is also considered valid.
（1）Example 1:
Input: “()”
Output: true
（2）Example 2:
Input: “()[]{}”
Output: true
（3）Example 3:
Input: “(]”
Output: false
（4）Example 4:
Input: “([)]”
Output: false
（5）Example 5:
Input: “{[]}”
Output: true
“”"
代码：
class Solution:
def isValid(self, s) :
stack = []
lookup = {"(": ")", "{": "}", "[": "]"}
for parenthese in s:
if parenthese in lookup:
stack.append(parenthese)
elif len(stack) == 0 or lookup[stack.pop()] != parenthese:
return False
return len(stack) == 0
注意要考虑特殊情况：（1）“]” 就是只有一个右边的括号，是错的。此时第八行的len(stack）== 0 （2）“[” 只有一个左边的括号也是错的，放在最后来判断。就是 return len(stack) == 0时候返回false
7.第21题（简单） Merge Two Sorted Lists 合并两个已经排好序的列表
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.
Example:
Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4
主要思想:
新建一个链表，然后比较两个链表中的元素值，把较小的那个链到新链表中，
由于两个输入链表的长度可能不同（但都是排好序的），
所以最终会有一个链表先完成插入所有元素，则直接另一个未完成的链表直接链入新链表的末尾。
代码：
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def mergeTwoLists(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
cur = dummy = ListNode(0) #创建一个头节点
# dummy：虚拟的node，一直指向原点。为了最后返回结果
while l1 and l2: # 遍历l1和l2
if l1.val < l2.val:
cur.next = l1
cur = cur.next
l1 = l1.next
else:
cur.next = l2
cur = cur.next
l2 = l2.next
cur.next = l1 or l2
return dummy.next
- 第26题 （简单）Remove Duplicates from Sorted Array 从有序数组中去除相同的元素
Given a sorted array nums,
remove the duplicates in-place such that each element appear only once and return the new length.
Do not allocate extra space for another array,
you must do this by modifying the input array in-place with O(1) extra memory.
（1）Example 1:
Given nums = [1,1,2],
Your function should return length = 2,
with the first two elements of nums being 1 and 2 respectively.
（2）Example 2:
Given nums = [0,0,1,1,1,2,2,3,3,4],
Your function should return length = 5,
with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.
该题的主要思想是，是要设立一个游标指向原点，然后循环，遇到相同的数时游标就不动，不同的数时游标就前进一位，并赋予当前这个不同的数。
代码：
class Solution:
def removeDuplicates(self, nums) :
if not nums:
return 0
cur = 0 # 设置一个游标指向原点，但遇到不同的数时，才移动
for i in range(len(nums)):
if nums[cur] != nums[i]:
cur += 1
nums[cur] = nums[i]
return cur + 1
9.第27题 （简单） Remove Element 移动元素
Given an array nums and a value val,
remove all instances of that value in-place and return the new length.
（1）Example 1:
Given nums = [3,2,2,3], val = 3,
your function should return length = 2, with the first two elements of nums being 2.
（2）Example 2:
Given nums = [0,1,2,2,3,0,4,2], val = 2,
your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.
思想：运用栈的思想，将数组看成一个栈，依次从后往前比较，若发现目标值将该元素出栈即可。
代码：
class Solution:
def removeElement(self, nums, val)
n = len(nums)
for i in range(n-1,-1,-1): # 逆序遍历（出栈是最后一个元素先出）
if nums[i] == val: # 与目标值相等
nums.pop(i) # 将该元素出栈
return len(nums) # 返回更新后数组的长度
10.第28题 Implement strStr() 简单题
Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.
（1）Example 1:
Input: haystack = “hello”, needle = “ll”
Output: 2
（2）Example 2:
Input: haystack = “aaaaa”, needle = “bba”
Output: -1
代码：
(1)
class Solution:
def strStr(self, haystack, needle):
for i in range(len(haystack) - len(needle)+1): # 并不是遍历全部
if haystack[i:i+len(needle)] == needle:
return i
return -1
(2) 用到自带函数，index… 不过也要掌握上面方法
class Solution(object):
def strStr(self, haystack, needle):
if needle in haystack:
return haystack.index(needle)
else:
return -1