[Q1] Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,Because nums[0] + nums[1] = 2 + 7 = 9,return [0, 1]. Solution

class Solution(object):    def twoSum(self, nums, target):        """        :type nums: List[int]        :type target: int        :rtype: List[int]        """        idx = {}        for l,x in enumerate(nums):            if target - x in idx: return [idx[target-x], l] idx[x]=l

问题：可能存在两个数相同的情况，而index（）函数对于重复元素只可返回第一个下标。

解决：使用字典

 

[Q2] 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 -> 8Explanation: 342 + 465 = 807. Solutions: https://leetcode.com/problems/add-two-numbers/discuss/1016/Clear-python-code-straight-forward

# Definition for singly-linked list.# class ListNode:#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution:    def addTwoNumbers(self, l1, l2):        """        :type l1: ListNode        :type l2: ListNode        :rtype: ListNode        """        carry = 0        l = l3 = ListNode(0)        while l1 or l2 or carry: cur1 = cur2 = 0 # avoid when l1 goes to end before l2 if l1: cur1 = l1.val # current value l1 = l1.next # go to next node if l2: cur2 = l2.val l2 = l2.next carry,cur3 = divmod(cur1+cur2+carry,10) l3.next = ListNode(cur3) l3 = l3.next return l.next

 

[Q3] Given a string, find the length of the longest substring without repeating characters.

Example 1:

Input: "abcabcbb"Output: 3 Explanation: The answer is "abc", with the length of 3.

Example 2:

Input: "bbbbb"Output: 1Explanation: The answer is "b", with the length of 1.

Example 3:

Input: "pwwkew"Output: 3Explanation: The answer is "wke", with the length of 3.              Note that the answer must be a substring, "pwke" is a subsequence and not a substring. Solution:  if s[j]s[j] have a duplicate in the range [i, j)[i,j) with index j'j′, we don't need to increase ii little by little. We can skip all the elements in the range [i, j'][i,j′] and let ii to be j' + 1j′+1 directly. 创建左节点i和右节点j，扫描窗口，若s【i:j】内有与s【j】相同的数，且该数位置为j`，则直接将i跳至j`，而j+1

class Solution:    def lengthOfLongestSubstring(self, s):        """        :type s: str        :rtype: int        """        i,j = 0,1        maxL,L = 0,0    # window length        if(len(s)<=1):            return len(s)        while i