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描述
You are given a string s consisting of digits and an integer k.
A round can be completed if the length of s is greater than k. In one round, do the following:
- Divide s into consecutive groups of size k such that the first k characters are in the first group, the next k characters are in the second group, and so on. Note that the size of the last group can be smaller than k.
- Replace each group of s with a string representing the sum of all its digits. For example, "346" is replaced with "13" because 3 + 4 + 6 = 13.
- Merge consecutive groups together to form a new string. If the length of the string is greater than k, repeat from step 1.
Return s after all rounds have been completed.
Example 1:
Input: s = "11111222223", k = 3
Output: "135"
Explanation:
- For the first round, we divide s into groups of size 3: "111", "112", "222", and "23".
Then we calculate the digit sum of each group: 1 + 1 + 1 = 3, 1 + 1 + 2 = 4, 2 + 2 + 2 = 6, and 2 + 3 = 5.
So, s becomes "3" + "4" + "6" + "5" = "3465" after the first round.
- For the second round, we divide s into "346" and "5".
Then we calculate the digit sum of each group: 3 + 4 + 6 = 13, 5 = 5.
So, s becomes "13" + "5" = "135" after second round.
Now, s.length <= k, so we return "135" as the answer.
Example 2:
Input: s = "00000000", k = 3
Output: "000"
Explanation:
We divide s into "000", "000", and "00".
Then we calculate the digit sum of each group: 0 + 0 + 0 = 0, 0 + 0 + 0 = 0, and 0 + 0 = 0.
s becomes "0" + "0" + "0" = "000", whose length is equal to k, so we return "000".
Note:
1 <= s.length <= 100
2 <= k <= 100
s consists of digits only.
解析
根据题意，给定一个由数字组成的字符串 s 和整数 k 。如果 s 的长度大于 k ，则可以完成一轮操作。 在一轮操作中，执行以下操作：
- 将 s 分成大小为 k 的连续数字组，使得前 k 个字符在第一组中，接下来的 k 个字符在第二组中，依此类推。 请注意，最后一组的长度可能小于 k 。
- 将每组 s 替换为表示其所有数字之和的字符串。 例如，“346”被替换为“13”，因为 3 + 4 + 6 = 13。
- 将连续的组合并在一起形成一个新的字符串。 如果字符串的长度大于 k ，则从步骤 1 开始重复新一轮的操作。
最后完成后返回 s 。
这道题考察的就是字符串和列表的基本操作，我们的解题也可以按照题目的要求进行：
- 如果 s 的长度小于等于 k 直接返回 s
- 否则进行 while 循环，将 s 拆分成若干 k 长度的子字符串，然后将每个子字符串进行相加变为新的子字符串放入列表 tmp 中，将 tmp 拼接成字符串赋与 s ，进行下一次循环操作
- 循环结束直接返回 s 即可
时间复杂度为 O(N)，空间复杂度为 O(N)。
解答
class Solution(object):
def digitSum(self, s, k):
"""
:type s: str
:type k: int
:rtype: str
"""
if len(s) <=k : return s
while len(s)>k:
tmp = []
for i in range(0, len(s), k):
tmp.append(s[i:i+k])
for i,t in enumerate(tmp):
sum = 0
for j in tmp[i]:
sum += int(j)
tmp[i] = str(sum)
s = ''.join(tmp)
return s
运行结果
121 / 121 test cases passed.
Status: Accepted
Runtime: 16 ms
Memory Usage: 13.7 MB
原题链接
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