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.
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| 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"
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Notes: calculate one mega shift. Do shift with two substring method calls.
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| class Solution {
public String stringShift(String s, int[][] shift) {
int n = s.length();
int rightShift = 0;
for (int i = 0; i < shift.length; i++) {
int[] one = shift[i];
if (one[0] == 0) {
rightShift = rightShift + (n - one[1]);
} else {
rightShift = rightShift + one[1];
}
}
rightShift = rightShift % n;
return s.substring(n - rightShift) + s.substring(0, n - rightShift);
}
}
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