[Leetcode] Count and Say
2014-07-27 16:03
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问题:The count-and-say sequence is the sequence of integers beginning as follows:
Given an integer n, generate the nth sequence.
Note: The sequence of integers will be represented as a string.
题目比较难懂,就是要输出上面第一行的序列,输入参数n表示要输出前n个。这行序列的意思是数上个字符串中的数值个数,n=1时输出字符串1;n=2时,因为上次字符串是1,有1个1,所以输出11;n=3时,由于上次字符是11,有2个1,所以输出21;n=4时,由于上次字符串是21,有1个2和1个1,所以输出1211,一次类推。用一个比较简单的迭代即可完成。
string itercont(string str)
{
string res;
int cont=1;
int i=0;
for( i=0; i<str.size()-1; i++)
{
if(str[i]==str[i+1])
cont++;
else
{
res.push_back('0'+cont);
res.push_back(str[i]);
cont=1;
}
}
if(i==str.size()-1)
{
res.push_back('0'+cont);
res.push_back(str[i]);
}
return res;
}
string countAndSay(int n)
{
if (n <= 0) return string();
string result="1";
for(int i=1; i<n; i++)
{
result=itercont(result);
}
return result;
}
1, 11, 21, 1211, 111221, ...
1is read off as
"one 1"or
11.
11is read off as
"two 1s"or
21.
21is read off as
"one 2, then
one 1"or
1211.
Given an integer n, generate the nth sequence.
Note: The sequence of integers will be represented as a string.
题目比较难懂,就是要输出上面第一行的序列,输入参数n表示要输出前n个。这行序列的意思是数上个字符串中的数值个数,n=1时输出字符串1;n=2时,因为上次字符串是1,有1个1,所以输出11;n=3时,由于上次字符是11,有2个1,所以输出21;n=4时,由于上次字符串是21,有1个2和1个1,所以输出1211,一次类推。用一个比较简单的迭代即可完成。
string itercont(string str)
{
string res;
int cont=1;
int i=0;
for( i=0; i<str.size()-1; i++)
{
if(str[i]==str[i+1])
cont++;
else
{
res.push_back('0'+cont);
res.push_back(str[i]);
cont=1;
}
}
if(i==str.size()-1)
{
res.push_back('0'+cont);
res.push_back(str[i]);
}
return res;
}
string countAndSay(int n)
{
if (n <= 0) return string();
string result="1";
for(int i=1; i<n; i++)
{
result=itercont(result);
}
return result;
}
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