# 【LeetCode】 Perform String Shifts

## Description
> 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.

> 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`

> 給你一個字串`s`包含小寫英文字母，和一個矩陣 `shift`，結構為`shift[i] = [方向, 大小]`：
> * `方向`可以是`0`(往左移)或是`1`(往右移)。
> * `大小`代表字串`s`要位移的量。
> * 往左移1代表去掉`s`的第一個字母並把它加到最後面。
> * 以此類推，往右移1代表去掉`s`的最後一個字母並把他加到最前面。

> 回傳最後運算完的字串。

> 限制：
> * `1 <= s.length <= 100`
> * `s` 只會包含小寫英文字母。
> * `1 <= shift.length <= 100`
> * `shift[i].length == 2`
> * `0 <= shift[i][0] <= 1`
> * `0 <= shift[i][1] <= 100`

## Example:

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"

## Solution

* 移過去在移回來可以抵銷，因此不要傻傻地每個`shift`都移動，請先算完最後要移動多少，動一次就好。
* 位移的部分，我直接根據原字串重建一個新的字串。
* 先找到第一個應該加入的字母，然後用`%`加一圈回來。
* 中間使用`while`讓`count`必定為正數。

### Code

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

###### tags: `LeetCode` `C++`