1427.Perform String Shifts

Tags: String Easy

Links: https://leetcode.com/explore/challenge/card/30-day-leetcoding-challenge/529/week-2/3299/

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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

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class Solution {
public:
    string stringShift(string s, vector<vector<int>>& shift) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        int sum = 0, len = shift.size();
        for (int i = 0; i < len; ++i) {
            if (shift[i][0] == 0) sum -= shift[i][1];
            else sum += shift[i][1];
        }

        int n = s.size();
        while (sum >= n) sum -= n;
        while (sum < 0) sum += n;
        sum = n - sum;

        if (sum == 0) return s; //放置出现循环的情况

        int d = GCD(sum, n);
        for (int i = 0; i < d; ++i) {
            int pos = i;
            char ch = s[i];
            while (true) {
                int j = pos + sum;
                if (j >= n) j -= n;
                if (j == i) break;
                s[pos] = s[j];
                pos = j;
            }
            s[pos] = ch;
        }

        return s;
    }

    int GCD(int a, int b)
    {
        return b == 0 ? a : GCD(b, a % b);
    }
};

数组移位技巧在字符串中的应用。