Day2 -- Shifting Letters

From: https://leetcode.com/

mode: random pick

degree: Medium

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

My solution:

 1 #!/usr/bin/env python
 2 # -*- coding: utf-8 -*-
 3 # Time    : 2018/11/8 
 4 
 5 
 6 class Solution:
 7     def __init__(self):
 8         self.s_list = []
 9         self.result = []
10 
11     def shiftingLetters(self, S, shifts):
12         """
13         :type S: str
14         :type shifts: List[int]
15         :rtype: str
16         """
17         self.s_list = [ord(i) for i in list(S)]
18 
19         def temp(x):
20             req = [self.plusIndex(i, x) for i in self.s_list]
21             self.s_list = req
22         map(temp, shifts)
23         self.result = [chr(i) for i in self.s_list]
24 
25     def plusIndex(self, index_node, times, length=26, start_index=97, end_index=122):
26         times_y = times % length
27         req_index = times_y + index_node
28         if req_index > end_index:
29             req_index = req_index - end_index + start_index - 1
30         return req_index
31 
32 
33 if __name__ == '__main__':
34     s = Solution()
35     s.shiftingLetters("dfsr", [1, 3])
36     print s.result