leetcode1427 - Perform String Shifts - easy

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"

 

先计算net shifts,  再进行string的分割和拼接。N = array shift的长度， L = string s的长度，time complexity O(N+L) (step 1 traverse the array and step 2 perform a single string-shift operation)

注意substr的用法：

substr(pos) - 从index为pos(包含)直到end

substr(pos, len) - 从index为pos取len位

 

实现：

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