[LeetCode] 1427. 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.

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

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"

题意是给一个string S和一个二维数组shift，请你根据shift里面的情况改动S，最后输出S被修改过后的结果。思路是首先遍历shift，判断出最后S到底是往左shift还是往右shift，用一个变量count记录。同时需要注意的是如果这个count的长度大于S的长度，需要取模。最后用substring函数看到底是S的两个substring到底应该如何拼接。

时间O(n)

空间O(1)

Java实现

class Solution {
    public String stringShift(String s, int[][] shift) {
        int count = 0;
        for (int[] value : shift) {
            if (value[0] == 0) {
                count += value[1];
            } else {
                count -= value[1];
            }
        }
        count = count % s.length();
        if (count > 0) {
            s = leftShift(s, count);
        } else if (count < 0) {
            s = rightShift(s, -count);
        }
        return s;
    }
    
    public static String leftShift(String s, int num) {
        return s.substring(num) + s.substring(0, num);
    }
    
    public static String rightShift(String s, int num) {
        return s.substring(s.length() - num) + s.substring(0, s.length() - num);
    }
}

LeetCode 题目总结