【LeetCode算法题库】Day7：Remove Nth Node From End of List & Valid Parentheses & Merge Two Lists

【Q19】

Given a linked list, remove the n-th node from the end of list and return its head.

Example:

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:

Given n will always be valid.

Follow up:

Could you do this in one pass?

# Definition for singly-linked list.

# class ListNode:

#     def __init__(self, x):

#         self.val = x

#         self.next = None

class Solution:

    def removeNthFromEnd(self, head: 'ListNode', n: 'int') -> 'ListNode':

        cur = head

        length = 0

        while cur!=None:

            cur = cur.next

            length += 1

        if length==0:

            return head

        else:

            idx = 0

            if length-n==0:

                return head.next

            else:

                cur = head

                while idx<length-n-1:

                    cur = cur.next

                    idx += 1

                cur.next = cur.next.next

                return head

【Q20】

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.

Note that an empty string is also considered valid.

Example 1:

Input: "()"

Output: true

Example 2:

Input: "()[]{}"

Output: true

Example 3:

Input: "(]"

Output: false

Example 4:

Input: "([)]"

Output: false

Example 5:

Input: "{[]}"

Output: true

解法：用堆栈。遍历字符串数组，把左括号全部压栈，遇到右括号时，判断与栈顶的左括号是否为一对，若是，则令栈顶的左括号出栈，判断遍历完毕的栈是否为空。若是，则返回True，否则返回False。
详解（直接看最后的solution）：https://leetcode.com/problems/valid-parentheses/solution/

class Solution:

    def isValid(self, s: 'str') -> 'bool':

        charmap = {')':'(',']':'[','}':'{'}

        if s==None:

            return True

        if len(s)%2!=0:

            return False

        stack = []

        for i in range(len(s)):

            if i==0:

                stack.append(s[i])

            elif s[i] in charmap:

                c = stack.pop()

                if c!=charmap.get(s[i]):

                    return False

            else:

                stack.append(s[i])

        return not stack

【Q21】

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

注意保存链表头！

# Definition for singly-linked list.

# class ListNode:

#     def __init__(self, x):

#         self.val = x

#         self.next = None

class Solution:

    def mergeTwoLists(self, l1: 'ListNode', l2: 'ListNode') -> 'ListNode':

        head = l = ListNode(None)

        while l1 and l2:

            if l1.val<l2.val:

                l.next = l1

                l1 = l1.next

            else:

                l.next = l2

                l2 = l2.next

            l = l.next

        if not l1:

            l.next = l2

        else:

            l.next = l1

        return head.next

巴特西

【LeetCode算法题库】Day7：Remove Nth Node From End of List & Valid Parentheses & Merge Two Lists

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