腾讯云短信正文模板每个变量取值最多支持6个字符出现的问题及应对方法

立个flag，1-100题每天分配10题，不会就先空着（7）。

1. 11：盛最多水的容器

class Solution:
    def maxArea(self, height: List[int]) -> int:
        res = 0
        left = 0
        right = len(height) - 1
        while left < right:
            area = (right - left) * min(height[left], height[right])
            res = area if area > res else res
            if height[left] < height[right]:
                left = left + 1
            else:
                right = right - 1
        return res

2. 12：整数转罗马数字

class Solution:
    def intToRoman(self, num: int) -> str:
        res = ""
        haxi = {
            1000: "M", 900: "CM", 500: "D", 400: "CD", 100: "C",
            90: "XC", 50: "L", 40: "XL", 10: "X",
            9: "IX", 5: "V", 4: "IV", 1: "I"
        }
        for key in haxi:
            while key <= num:
                num = num - key
                res = res + haxi[key]
        return res

3. 13：罗马数字转整数 

class Solution:
    def romanToInt(self, s: str) -> int:
        res = 0
        haxi = {"I": 1, "V": 5, "X": 10, "L": 50, "C": 100, "D": 500, "M": 1000}
        for i in range(len(s) - 1):
            if haxi[s[i]] >= haxi[s[i + 1]]:
                res = res + haxi[s[i]]
            else:
                res = res - haxi[s[i]]
        res = res + haxi[s[-1]]
        return res

4. 14：最长公共前缀

class Solution:
    def longestCommonPrefix(self, strs: List[str]) -> str:
        def get_same(str1, str2):
            res = ''
            i = 0
            while i < len(str1) and i < len(str2):
                if str1[i] == str2[i]:
                    res = res + str1[i]
                    i = i + 1
                else:
                    break
            return res
        
        res = strs[0]
        for i in range(len(strs)):
            res = get_same(res, strs[i])
        return res

 5. 15：三数之和

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        res = []
        nums.sort()
        for i in range(len(nums) - 2):
            if i > 0 and nums[i] == nums[i - 1]:
                continue
            left = i + 1
            right = len(nums) - 1
            while left < right:
                sum = nums[i] + nums[left] + nums[right]
                if sum > 0:
                    right = right - 1
                elif sum < 0:
                    left = left + 1
                elif sum == 0:
                    res.append([nums[i], nums[left], nums[right]])
                    left = left + 1
                    while left < right and nums[left] == nums[left - 1]:
                        left = left + 1
                    right = right - 1
                    while left < right and nums[right] == nums[right + 1]:
                        right = right - 1
        return res

 6. 16：最接近的三数之和

class Solution:
    def threeSumClosest(self, nums: List[int], target: int) -> int:
        minimum = float("inf")
        nums.sort()
        for i in range(len(nums) - 2):
            if i > 0 and nums[i] == nums[i - 1]:
                continue
            left = i + 1
            right = len(nums) - 1
            while left < right:
                sum = nums[i] + nums[left] + nums[right]
                if sum == target:
                    return sum
                elif sum < target:
                    left = left + 1
                elif sum > target:
                    right = right - 1
                if abs(sum - target) < minimum:
                    res = sum
                    minimum = abs(sum - target)
        return res

 7. 17：电话号码的字母组合

 8. 18：四数之和

class Solution:
    def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
        res = []
        nums.sort()
        for i in range(len(nums) - 3):
            if i > 0 and nums[i] == nums[i - 1]:
                continue
            for j in range(i + 1, len(nums) - 2):
                if j > i + 1 and nums[j] == nums[j - 1]:
                    continue
                left = j + 1
                right = len(nums) - 1
                while left < right:
                    sum = nums[i] + nums[j] + nums[left] + nums[right]
                    if sum < target:
                        left = left + 1
                    elif sum > target:
                        right = right - 1
                    elif sum == target:
                        res.append([nums[i], nums[j], nums[left], nums[right]])
                        left = left + 1
                        while left < right and nums[left] == nums[left - 1]:
                            left = left + 1
                        right = right - 1
                        while left < right and nums[right] == nums[right + 1]:
                            right = right - 1
        return res

 9. 19：删除链表的倒数第N个结点

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
        dummy = slow = fast = ListNode()
        dummy.next = head
        while n:
            fast = fast.next
            n = n - 1
        while fast.next:
            slow = slow.next
            fast = fast.next
        slow.next = slow.next.next
        return dummy.next

10. 20：有效的括号 

class Solution:
    def isValid(self, s: str) -> bool:
        stack = []
        pair = {"(": ")", "[": "]", "{": "}"}
        for i in s:
            if i in "([{":
                stack.append(pair[i])
            else:
                if stack and stack[-1] == i:
                    stack.pop()
                else:
                    return False
        if stack:
            return False
        else:
            return True