【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++