字符串操作-分解字符串并补0

题目描述

按要求分解字符串：
输入两个数M,N
M：输入的M串字符串
N：输出的每串字符串的位数，不够补0。

eg
input：2 8
abc
123456789
out：abc00000
12345678
90000000

解题思路一

1. C++字符串常用函数:

• string substr (size_t pos = 0, size_t len = npos) const;
  note:第二个参数为要取的字符串长度

#include <iostream>
#include <vector>
#include <stack>
#include <string>
using namespace std;
class Solution {
public:
vector<string> splitStr(vector<string> inStrs, int n);
};
int main()
{
Solution sol;
vector<string> inStrs;
vector<string> outStrs;
int M = 0;
int N = 0;
cin >> M >> N;
for (int i = 0; i < M; i++)
{
string str;
cin >> str;
inStrs.push_back(str);
}
outStrs = sol.splitStr(inStrs, N);
for (int i = 0; i < outStrs.size(); i++)
{
cout << outStrs[i] << "
";
}
cout << endl;
system("pause");
return 0;
}
vector<string> Solution::splitStr(vector<string> inStrs, int n)
{
vector<string> outStrs;
for (int i = 0; i < inStrs.size(); i++)
{
string str = inStrs[i];
int strLen = str.length();
if (strLen <= n)//字符串长度小于n，补0
{
for (int j = 0; j < (n - strLen); j++)
{
str += "0";
}
outStrs.push_back(str);
}
else//字符串长度大于n
{
int subLen = strLen;
int beg = 0;
int end = n;
string strTmp;
while (subLen >= n)//多少个n
{
strTmp = str.substr(beg, n);
beg += n;
end += n;
outStrs.push_back(strTmp);
subLen -= n;
}
if (subLen > 0)//余下的补0
{
strTmp = str.substr(beg, strLen);
int addZeors = n - subLen;
for (; addZeors > 0; addZeors--)
{
strTmp += "0";
}
outStrs.push_back(strTmp);
}
}
}
return outStrs;
}

解题思路二

• 利用递归的思想

最后

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