9.6 Rotate String

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2023-12-01

Question

lintcode: Rotate String

Problem Statement

Given a string and an offset, rotate string by offset. (rotate from left to right)

Example

Given "abcdefg".

offset=0 => "abcdefg"
offset=1 => "gabcdef"
offset=2 => "fgabcde"
offset=3 => "efgabcd"

Challenge

Rotate in-place with O(1) extra memory.

题解

常见的翻转法应用题，仔细观察规律可知翻转的分割点在从数组末尾数起的offset位置。先翻转前半部分，随后翻转后半部分，最后整体翻转。

Python - immutable string

class Solution:
    """
    param A: A string
    param offset: Rotate string with offset.
    return: Rotated string.
    """
    def rotateString(self, A, offset):
        if A is None or len(A) == 0:
            return A

        offset %= len(A)
        before = A[:len(A) - offset]
        after = A[len(A) - offset:]
        # [::-1] means reverse in Python
        A = before[::-1] + after[::-1]
        A = A[::-1]

        return A

Python - mutable list

class Solution:
    # @param A: a list of char
    # @param offset: an integer
    # @return: nothing
    def rotateString(self, A, offset):
        if A is None or len(A) == 0:
            return

        offset %= len(A)
        self.reverse(A, 0, len(A)-offset-1)
        self.reverse(A, len(A)-offset, len(A)-1)
        self.reverse(A, 0, len(A)-1)

    def reverse(self, str_l, start, end):
        while start < end:
            str_l[start], str_l[end] = str_l[end], str_l[start]
            start += 1
            end -= 1

C++

class Solution {
public:
  /**
     * param A: A string
     * param offset: Rotate string with offset.
     * return: Rotated string.
     */
    string rotateString(string A, int offset) {
        if (A.empty() || A.size() == 0) {
            return A;
        }

        int len = A.size();
        offset %= len;
        reverse(A, 0, len - offset - 1);
        reverse(A, len - offset, len - 1);
        reverse(A, 0, len - 1);
        return A;
    }

private:
    void reverse(string &str, int start, int end) {
        while (start < end) {
            char temp = str[start];
            str[start] = str[end];
            str[end] = temp;
            start++;
            end--;
        }
    }
};

Java

public class Solution {
    /*
     * param A: A string
     * param offset: Rotate string with offset.
     * return: Rotated string.
     */
    public char[] rotateString(char[] A, int offset) {
        if (A == null || A.length == 0) {
            return A;
        }

        int len = A.length;
        offset %= len;
        reverse(A, 0, len - offset - 1);
        reverse(A, len - offset, len - 1);
        reverse(A, 0, len - 1);

        return A;
    }

    private void reverse(char[] str, int start, int end) {
        while (start < end) {
            char temp = str[start];
            str[start] = str[end];
            str[end] = temp;
            start++;
            end--;
        }
    }
};

源码分析

异常处理，A为空或者其长度为0
offset可能超出A的大小，应模len后再用
三步翻转法

Python 虽没有提供字符串的翻转，但用 slice 非常容易实现，非常 Pythonic!

通常来说，字符串在各种编程语言中的实现一般为 immutable 的，对字符串做改变时往往会生成新的字符串，所以如果要达到空间复杂度为 O(1) 的效果，需要用可变数据结构来实现。

复杂度分析

翻转一次时间复杂度近似为 $$O(n)$$, 原地交换的空间复杂度为 $$O(1)$$, 非原地交换的空间复杂度为 $$O(n)$$. 总共翻转3次，所以总的时间复杂度为 $$O(n)$$, 空间复杂度为 $$O(1)$$ 或者 $$O(n)$$.