leetcode Perform String Shifts

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

题目大意：给定一个只包含小写字母的字符串s和一个转移(shift）矩阵，其中shift[i] = [direction, amount]：

direction只能是0或者1，0表示字符串左移，1表示右移，

amount表示字符串s的移动的数量

左移1次表示将s的第一个字符移除，并将其插入(append)到字符串尾

同样的，右移1次表示将s的最后一个字符移除，并将其插入(append)到字符串首

最后返回经过所有操作后的串。

思路：假设字符串长度为len，首先明确几个规律：

1）右移或者左移len位相当于不移动

2）左移num位相当于右移len-num位

3）左移num位再右移num位相当于原字符串不移动，所以本题可以一次性统计，算出最后右移的位数（左移的位数用右移来抵消）

4）右移num位，常规做法就是模拟，一位一位右移，计算复杂度为O(num * len), 一种做法是翻转字符串前len-num个字符，翻转最后num个字符，最后再翻转整个字符，复杂度为O(len):reverse(s[:len-num]), reverse(s[len-num:], reverse(s)

C++代码：

 1 class Solution {
 2 public:
 3     string stringShift(string s, vector<vector<int>>& shift) {
 4         int shift_right = 0; //shift_right记录总的右移位数
 5         string ans(s);
 6         for (auto shift_i : shift) {
 7             if (shift_i[0] == 0) //左移，shift_right来抵消左移，用减法
 8                 shift_right -= shift_i[1]; 
 9             else
10                 shift_right += shift_i[1];
11         }
12         
13         int len = ans.length();
14         //shift_right = (shift_right - shift_right / len * len + len) % len; //此语句同时考虑shift_right正负两种情况，相当于下面两句
15         shift_right = shift_right % len; //先消除len的整数倍的右移次数
16         shift_right = shift_right < 0 ? shift_right + len : shift_right; //如果右移次数为负，转换成真正的右移
17         int mid = len - shift_right;
18         
19         for (int i = 0, j = mid - 1; i < j; ++i, --j) //翻转后半部分字符串
20             swap(ans[i], ans[j]);
21         for (int i = mid, j = len - 1; i < j; ++i, --j) //翻转前半部分字符串
22             swap(ans[i], ans[j]);
23         for (int i = 0, j = len - 1; i < j; ++i, --j) //翻转整个字符串
24             swap(ans[i], ans[j]);
25         return ans;
26     }
27 };

python3代码

 1 class Solution:
 2     def stringShift(self, s: str, shift: List[List[int]]) -> str:
 3         shift_right = 0
 4         for x, y in shift:
 5             if x == 0:
 6                 shift_right -= y
 7             else:
 8                 shift_right += y
 9         shift_right %= len(s)
10         return s[-shift_right:] + s[:-shift_right]

标签：右移,right,Perform,shift,len,Shifts,num,ans,leetcode
来源： https://www.cnblogs.com/qinduanyinghua/p/12724971.html