LeetCode 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.

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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给定一个字符串s，并且给定一些向左或向右shift的操作，问经过这些操作之后，字符串最终的形式是怎样的。

简单题，虽然有很多操作，但可以先把向左的操作数累加，向右的操作数累加，最后比比大小，看最终是向左还是向右。完整代码如下：

class Solution {
public:
	string stringShift(string s, vector<vector<int>>& shift) {
		int left = 0, right = 0, n = s.size();
		for (int i = 0; i < shift.size(); ++i) {
			if (shift[i][0] == 0)left += shift[i][1];
			else right += shift[i][1];
		}
		if (left > right) {
			left -= right;
			left %= n;
			return s.substr(left) + s.substr(0, left);
		}
		else if (right > left) {
			right -= left;
			right %= n;
			return s.substr(n - right) + s.substr(0, n - right);
		}
		else {
			return s;
		}
	}
};

本代码提交AC，用时0MS。