LeetCode-Reverse_String_II

題目：

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]

翻譯：

給定一個字符串和一個數字k，你需要翻轉從字符串開始起每2k的字符的前k的字符。如果剩下的字符少於k個，那麼將它們都翻轉。如果剩下的字符少於2k但是大於等於k個，那麼翻轉前k個字符，剩下的保持不變。

例子:

輸入: s = "abcdefg", k = 2
輸出: "bacdfeg"

約束條件:

1. 字符串中只包含小寫的英文字母。
2. 給定的字符串和k的長度在[1,10000]範圍內。
   

思路：

從頭開始掃描所有2k的子字符串，根據條件進行判斷。

C++代碼(Visual  studio 2107):

#include "stdafx.h"
#include <iostream>
#include <string>
using namespace std;

class Solution {
public:
	string reverseStr(string s, int k) {
		for (int i = 0; i < s.size(); i += 2 * k) {
			if (i + k >= s.size())
				reverse(s.begin() + i, s.end());
			else
				reverse(s.begin() + i, s.begin()+i+k);
		}
		return s;
	}
};

int main()
{
	Solution s;
	string str = "abcdefg";
	int k = 2;
	string result = s.reverseStr(str, k);
	cout << result<<endl;
    return 0;
}