# day_14: Perform String Shifts

###### tags: `online_judge`, `python3`

給定一個字串和一連串的 round shift 操作，round shift 格式為```[方向，距離]```。求完成 round shift 操作後的字串。

round shift 操作會互相抵消，所以整合所有操作，實際上 shift 一次就可以了。

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## 題目

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

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## 解答

class Solution:
    def stringShift(self, s, shift):
        
        count = 0
        
        for op in shift:
            
            if op[0] == 0:
                count += op[1]
            else:
                count -= op[1]
            
        count %= len(s)
        
        s = s[count:] + s[:count]
        
        return s

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### 成績


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