LeetCode 30 days challenge: Perform String Shifts

題目名稱 Perform String Shifts

題目描述：
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.

例子：

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

我的理解
向左向右移動是一個互斥可逆的過程，因而可以相互抵消。所以遇到向左就加上向左的步數，遇到向右就減去向右的步數。最後統計向左的步數是多少，直接一步向左移動就可以了。當然向左的步數可能是負值，那就向右相反數步數就可以了。

C++ solution

class Solution {
public:
    string stringShift(string s, vector<vector<int>>& shift) {
        
        int lefshif = 0;
        
        for(int i = 0; i < shift.size(); i++){
            
            lefshif = (shift[i][0] == 0) ? lefshif + shift[i][1] : lefshif - shift[i][1];
            
        }
        
        if(lefshif > 0){
            lefshif = lefshif % s.length();
            s = s.substr(lefshif) + s.substr(0, lefshif);
        }
        else if(lefshif < 0){
            lefshif = (-lefshif) % s.length();
            
            s = s.substr(s.length() - lefshif) + s.substr(0, s.length()-lefshif);
        }
        return s;
    }
};