LintCode 1790 Rotate String II - Easy

1790. Rotate String II — Easy

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Problem

1790. Rotate String II

Given a string(Given in the way of char array), a right offset and a left offset, rotate the string by offset in place.(left offest represents the offset of a string to the left,right offest represents the offset of a string to the right,the total offset is calculated from the left offset and the right offset,split two strings at the total offset and swap positions)。

Example Example 1:

Input：str ="abcdefg", left = 3, right = 1 Output："cdefgab" Explanation：The left offset is 3, the right offset is 1, and the total offset is left 2. Therefore, the original string moves to the left and becomes "cdefg"+ "ab". Example 2:

Input：str="abcdefg", left = 0, right = 0 Output："abcdefg" Explanation：The left offset is 0, the right offset is 0, and the total offset is 0. So the string remains unchanged. Example 3:

Input：str = "abcdefg",left = 1, right = 2 Output："gabcdef" Explanation：The left offset is 1, the right offset is 2, and the total offset is right 1. Therefore, the original string moves to the left and becomes "g" + "abcdef".

Solution

### String 类型题目，注意越界(需要取模的操作)和slice下标问题。
 
class Solution:
    """
    @param str: A String
    @param left: a left offset
    @param right: a right offset
    @return: return a rotate string
    """
    def RotateString2(self, str, left, right):
        # write your code here
        total = left * (- 1) + right
        shift = abs(total)
        n = len(str)
        if total >= 0:
            left = str[(n - shift) % (n):]
            right = str[:(n - shift) % (n)]
        else:
            left = str[shift % n:]
            right = str[:shift % n]
        return left + right