shifting letters medium

String/ShiftingLetters.py

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+"""
+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
+
+
+给一个字符串，不断经过转换，得出最终字符串。
+
+每一轮的转换都是对这之前的所有字符串而言的。
+
+思路：
+1.
+ 从后向前，得到索引 % 26.
+
+2. 
+ 转换字符到 ascii，相加然后处理超过的量。
+
+测试地址：
+https://leetcode.com/contest/weekly-contest-88/problems/shifting-letters/
+
+"""
+class Solution(object):
+    def shiftingLetters(self, S, shifts):
+        """
+        :type S: str
+        :type shifts: List[int]
+        :rtype: str
+        """
+
+        letters = "abcdefghijklmnopqrstuvwxyz"
+
+        new_s = []
+        old_index = 0
+        for i, j in zip(S[::-1], shifts[::-1]):
+            index = letters.index(i)
+
+            new_index = (index+j+old_index)%26
+
+            new_s.append(letters[new_index])
+            old_index += j
+        return ''.join(new_s[::-1])
+