optimize code structure

Leetcode/00001_two_sum.py → Leetcode/1-50/00001_two_sum.py

@@ -1,4 +1,5 @@
 """
+https://leetcode.com/problems/two-sum/
 Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.
 
 You may assume that each input would have exactly one solution, and you may not use the same element twice.

Leetcode/1-50/00002_add_two_numbers.py

@@ -0,0 +1,92 @@
+"""
+https://leetcode.com/problems/add-two-numbers/
+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 contains a single digit. Add the two numbers and return the sum as a linked list.
+
+You may assume the two numbers do not contain any leading zero, except the number 0 itself.
+
+Example 1:
+
+
+Input: l1 = [2,4,3], l2 = [5,6,4]
+Output: [7,0,8]
+Explanation: 342 + 465 = 807.
+Example 2:
+
+Input: l1 = [0], l2 = [0]
+Output: [0]
+Example 3:
+
+Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
+Output: [8,9,9,9,0,0,0,1]
+
+
+Constraints:
+
+The number of nodes in each linked list is in the range [1, 100].
+0 <= Node.val <= 9
+It is guaranteed that the list represents a number that does not have leading zeros.
+"""
+from typing import Optional
+
+class ListNode:
+    def __init__(self, val=0, next=None):
+        self.val = val
+        self.next = next
+
+
+def addTwoNumbers(l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
+    l1_str = ""
+    while True:
+        l1_str += str(l1.val)
+        if l1.next is None:
+            break
+        l1 = l1.next
+    print(f"l1_str: {l1_str}")
+
+    l2_str = ""
+    while True:
+        l2_str += str(l2.val)
+        if l2.next is None:
+            break
+        l2 = l2.next
+
+    print(f"l2_str: {l2_str}")
+
+    big, small = l1_str, l2_str
+    if len(l1_str) < len(l2_str):
+        big, small = l2_str, l1_str
+
+    total = int(big) + int(small)
+    print(total)
+
+    node_list = list()
+    for s in str(total):
+        node_list.append(ListNode(int(s)))
+
+    for i, n in enumerate(node_list):
+        if i + 1 <= len(node_list) - 1:
+            n.next = node_list[i+1]
+
+    return node_list[0]
+
+
+if __name__ == '__main__':
+
+    # test link table
+    a1 = ListNode(2)
+    b1 = ListNode(4)
+    c1 = ListNode(3)
+    a1.next = b1
+    b1.next = c1
+
+    a2 = ListNode(5)
+    b2 = ListNode(6)
+    c2 = ListNode(4)
+    d2 = ListNode(6)
+    a2.next = b2
+    b2.next = c2
+    c2.next = d2
+
+    addTwoNumbers(a1, a2)
+
+    # print(a.next.next.val)

Leetcode/00003.py → Leetcode/1-50/00003_longest_substrings.py

@@ -17,6 +17,33 @@
 输出: 3
 解释: 因为无重复字符的最长子串是 "wke"，所以其长度为 3。
      请注意，你的答案必须是 子串 的长度，"pwke" 是一个子序列，不是子串。
+
+Given a string s, find the length of the longest
+substring without repeating characters.
+
+
+Example 1:
+
+Input: s = "abcabcbb"
+Output: 3
+Explanation: The answer is "abc", with the length of 3.
+Example 2:
+
+Input: s = "bbbbb"
+Output: 1
+Explanation: The answer is "b", with the length of 1.
+Example 3:
+
+Input: s = "pwwkew"
+Output: 3
+Explanation: The answer is "wke", with the length of 3.
+Notice that the answer must be a substring, "pwke" is a subsequence and not a substring.
+
+
+Constraints:
+
+0 <= s.length <= 5 * 104
+s consists of English letters, digits, symbols and spaces.
 """
 
 import time