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

My solution:

#!/usr/bin/env python
# -*- coding: utf-8 -*-
# Time    : 2018/11/8 


class Solution:
    def __init__(self):
        self.s_list = []
        self.result = []

    def shiftingLetters(self, S, shifts):
        """
        :type S: str
        :type shifts: List[int]
        :rtype: str
        """
        self.s_list = [ord(i) for i in list(S)]

        def temp(x):
            req = [self.plusIndex(i, x) for i in self.s_list]
            self.s_list = req
        map(temp, shifts)
        self.result = [chr(i) for i in self.s_list]

    def plusIndex(self, index_node, times, length=26, start_index=97, end_index=122):
        times_y = times % length
        req_index = times_y + index_node
        if req_index > end_index:
            req_index = req_index - end_index + start_index - 1
        return req_index


if __name__ == '__main__':
    s = Solution()
    s.shiftingLetters("dfsr", [1, 3])
    print s.result