【LeetCode】848. Shifting Letters 解题报告(Python)

时间:2023-03-09 22:19:04
【LeetCode】848. Shifting Letters 解题报告(Python)

【LeetCode】848. Shifting Letters 解题报告(Python)

标签(空格分隔): LeetCode

作者: 负雪明烛
id: fuxuemingzhu
个人博客: http://fuxuemingzhu.me/


题目地址:https://leetcode.com/problems/shifting-letters/description/

题目描述:

We have a string S of lowercase letters, and an integer array shifts.

Call the shift of a letter, the next letter in the alphabet, (wrapping around so that ‘z’ becomes ‘a’).

For example, shift(‘a’) = ‘b’, shift(‘t’) = ‘u’, and shift(‘z’) = ‘a’.

Now for each shifts[i] = x, we want to shift the first i+1 letters of S, x times.

Return the final string after all such shifts to S are applied.

Example 1:

Input: S = "abc", shifts = [3,5,9]
Output: "rpl"
Explanation:
We start with "abc".
After shifting the first 1 letters of S by 3, we have "dbc".
After shifting the first 2 letters of S by 5, we have "igc".
After shifting the first 3 letters of S by 9, we have "rpl", the answer.

Note:

  1. 1 <= S.length = shifts.length <= 20000
  2. 0 <= shifts[i] <= 10 ^ 9

题目大意

(这个题本身简单,但是读懂题目很重要)

给出了一个字符串S,以及和这个字符串等长的数组shifts。定义了一个shift操作:把某个字符在字母表上移动某位(字母’z’再向右移得到’a’)。现在遍历shifts,每个操作都是把当前位数之前的所有字符移动shift位。求最后得到的字符串。

解题方法

坑还是挺明显的:需要把当前位数之前的所有字符串都去shift操作。看出题目给的字符串挺长的,如果普通的遍历,在每次遍历的时候再把之前所有shift过的再次shift,那么就会超时。

所以正确的做法是先求出每个字符串需要shift的次数。即对shifts进行位置之后的求和。得出要shift的位数之后,按照题目给的那种循环去操作就好了。

(应该没有人傻到真的去循环,而不是用求余操作吧233,逃……)

class Solution(object):
def shiftingLetters(self, S, shifts):
"""
:type S: str
:type shifts: List[int]
:rtype: str
"""
_len = len(S)
shifts_sum = sum(shifts)
shifts_real = []
for shift in shifts:
shifts_real.append(shifts_sum)
shifts_sum -= shift
def shift_map(string, shift_time):
shifted = ord(s) + (shift_time % 26)
return chr(shifted if shifted <= ord('z') else shifted - 26)
ans = ''
for i, s in enumerate(S):
ans += shift_map(s, shifts_real[i])
return ans

日期

2018 年 6 月 10 日 ———— 等了两天的腾讯比赛复赛B的数据集,结果人家在复赛刚开始就给了。。