Perform String Shifts

https://leetcode.com/explore/challenge/card/30-day-leetcoding-challenge/529/week-2/3299/

You are given a string s containing lowercase English letters, and a matrix shift, where shift[i] = [direction, amount]:

• direction can be 0 (for left shift) or 1 (for right shift).

• amount is the amount by which string s is to be shifted.

• A left shift by 1 means remove the first character of s and append it to the end.

• Similarly, a right shift by 1 means remove the last character of s and add it to the beginning.

Return the final string after all operations.

Example 1:

Input: s = "abc", shift = [[0,1],[1,2]]
Output: "cab"
Explanation: 
[0,1] means shift to left by 1. "abc" -> "bca"
[1,2] means shift to right by 2. "bca" -> "cab"

Example 2:

Input: s = "abcdefg", shift = [[1,1],[1,1],[0,2],[1,3]]
Output: "efgabcd"
Explanation:  
[1,1] means shift to right by 1. "abcdefg" -> "gabcdef"
[1,1] means shift to right by 1. "gabcdef" -> "fgabcde"
[0,2] means shift to left by 2. "fgabcde" -> "abcdefg"
[1,3] means shift to right by 3. "abcdefg" -> "efgabcd"

Constraints:

• 1 <= s.length <= 100

• s only contains lower case English letters.

• 1 <= shift.length <= 100

• shift[i].length == 2

• 0 <= shift[i][0] <= 1

• 0 <= shift[i][1] <= 100

给定字符串s和数组shift，shift每个元素代表向左或向右移动多少步，问按照shift操作s后的结果。向左shift是(i - d + N) % N, 向右是(i + d - N) % N，一加一减。因此将所有shift操作的结果累加并把最终效果统一看成向右移动，统计最终向右移动了多少步，所以累加时向右是+，向左是-。

class Solution:
    def stringShift(self, s: str, shift: List[List[int]]) -> str:
        mv = 0
        for sh in shift:
            mv += sh[1] if sh[0] == 1 else -sh[1]
        mv %= len(s)
        return s[-mv:] + s[:-mv]