leetcode 541. Reverse String II(easy)

Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from the start of the string. If there are less than k characters left, reverse all of them. If there are less than 2k but greater than or equal to k characters, then reverse the first k characters and left the other as original.

Example:

Input: s = "abcdefg", k = 2
Output: "bacdfeg"

Restrictions:

1. The string consists of lower English letters only.
2. Length of the given string and k will in the range [1, 10000]

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题目的含义是每隔k个位置就截取k这么长的字符串反转，那么这里主要涉及到奇数段和偶数段字符串。

class Solution {
public:
    string reverseS(string s)
    {
        string result;
        stack<char> temp;
        for(int i=0;i<s.length();i++)
        {
            temp.push(s[i]);
            
        }
        while(!temp.empty())
        {
            result += temp.top();
            temp.pop();
        }
        return result;
    }
    string reverseStr(string s, int k) {
        string result;
        int len = s.length();
        int seg = len/k;
        if(seg == 0) return reverseS(s.substr(0,len));
        for(int i=1;i<=seg;i++)
        {
            if(i%2 != 0)
               result += reverseS(s.substr((i-1)*k,k));
            else
               result += s.substr((i-1)*k,k);
        }
        if(len>seg*k)
        {
            if(seg%2 == 0)
               result += reverseS(s.substr(seg*k,len-seg*k));
            else
               result += s.substr(seg*k,len-seg*k);
        }
        return result;
    }
};