2019/4/16
subject：771. Jewels and Stones
solution：
class Solution:
def numJewelsInStones(self, J, S):
"""
:type J: str
:type S: str
:rtype: int
"""
return sum(S.count(i) for i in J)
tips：
python中，count() 方法用于统计某个元素在列表中出现的次数。
subject: 1021. Remove Outermost Parentheses
solution:
class Solution:
def removeOuterParentheses(self, S: str) -> str:
count = 0
prev = None
res = []
left_n = 0
right_n = 0
j = 0
for i in range(len(S)):
if S[i] == "(":
left_n += 1 # add 1 when left parenthesis
else:
right_n += 1 # add 1 when right
if left_n == right_n:
res.append(S[j + 1:i]) # add part when left_n = right_n
j = i + 1 # because next first item will be cut, i + 1 gives j += 2 total
return ''.join(res)
2019/4/17
subject: 709. To Lower Case
solution1 :
class Solution:
def toLowerCase(self, str: str) -> str:
for i in str:
if ord(i) in range(65,91):
new = ord(i) + 32
n = chr(new)
str = str.replace(i,n)
return str
solution 2:
class Solution(object):
def toLowerCase(self, str):
"""
:type str: str
:rtype: str
"""
stack = [ord(x) for x in str]
for i in range(len(stack)):
if stack[i] > 64 and stack[i] < 91:
stack[i] += 32
asw=[chr(x) for x in stack]
return ''.join(asw)
tips：
Python join() 方法用于将序列中的元素以指定的字符连接生成一个新的字符串。
附：ASCII码表
2019/4/22
subject: 2.Add two numbers
solution:
#
# @lc app=leetcode id=2 lang=python3
#
# [2] Add Two Numbers
#
# https://leetcode.com/problems/add-two-numbers/description/
#
# algorithms
# Medium (30.92%)
# Total Accepted: 831.3K
# Total Submissions: 2.7M
# Testcase Example: '[2,4,3]\n[5,6,4]'
#
# You are given two non-empty linked lists representing two non-negative
# integers. The digits are stored in reverse order and each of their nodes
# contain a single digit. Add the two numbers and return it as a linked list.
#
# You may assume the two numbers do not contain any leading zero, except the
# number 0 itself.
#
# Example:
#
#
# Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
# Output: 7 -> 0 -> 8
# Explanation: 342 + 465 = 807.
#
#
#
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
carry = 0
root = n = ListNode(0)
while l1 or l2 or carry:
v1 = v2 = 0
if l1:
v1 = l1.val
l1 = l1.next
if l2:
v2 = l2.val
l2 = l2.next
carry, val = divmod(v1+v2+carry, 10)
n.next = ListNode(val)
n = n.next
return root.next
tips:
这道题涉及到Python数据结构之链表（linked list）
root的作用：
2019/4/23
subject: [5] Longest Palindromic Substring
solution:
#
# @lc app=leetcode id=5 lang=python3
#
# [5] Longest Palindromic Substring
#
# https://leetcode.com/problems/longest-palindromic-substring/description/
#
# algorithms
# Medium (27.00%)
# Total Accepted: 528.9K
# Total Submissions: 2M
# Testcase Example: '"babad"'
#
# Given a string s, find the longest palindromic substring in s. You may assume
# that the maximum length of s is 1000.
#
# Example 1:
#
#
# Input: "babad"
# Output: "bab"
# Note: "aba" is also a valid answer.
#
#
# Example 2:
#
#
# Input: "cbbd"
# Output: "bb"
#
#
#
class Solution:
def longestPalindrome(self, s):
res = ""
for i in range(len(s)):
# odd case, like "aba"
tmp = self.helper(s, i, i)
if len(tmp) > len(res):
res = tmp
# even case, like "abba"
tmp = self.helper(s, i, i+1)
if len(tmp) > len(res):
res = tmp
return res
# get the longest palindrome, l, r are the middle indexes
# from inner to outer
def helper(self, s, l, r):
while l >= 0 and r < len(s) and s[l] == s[r]:
l -= 1
r += 1
return s[l+1:r]
tips:
这道题主要是需要理解palindromic的格式，即共有“a" "aa" "aba" 三种形式。另外可以学习到的是在solution类中再编写辅助函数调用的思想。
helper中的思想也值得注意，l -= 1 r += 1的操作是为了让扫描区域向str两边扩散。