【LeetCode算法题库】Day1：TwoSums & Add Two Numbers & Longest Substring Without Repeating Characters

[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 -> 8
Explanation: 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 = None

class 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: 1
Explanation: The answer is "b", with the length of 1.

Example 3:

Input: "pwwkew"
Output: 3
Explanation: 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<len(s) and j<len(s):
            if s[j] not in s[i:j]:
                j = j+1
            else:
                i = i+1
            maxL = max(maxL,j-i)
        return maxL