[leetcode] 【字符串】 38. Count and Say
2016-06-15 15:53
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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.
11的下一个数是2个1 表示为21
21的下一个数为1个2 1个1 表示为1211
以此类推,n表示第n个数。
class Solution {
public:
string tran(string cur)
{
string res;
char key=cur[0];
int times=1;
for(int i=1;i!=cur.size();i++)
{
if(cur[i]==key)
times++;
else
{
char stimes=times+'0';
res=res+stimes+key;
key=cur[i];
times=1;
}
}
char stimes=times+'0';
res=res+stimes+key;
return res;
}
string countAndSay(int n) {
string cur="1";
while(--n)
cur=tran(cur);
return cur;
}
};
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.
题意
按照计数方式来生成下一个数值,比如1的下一个数值就是1个1 表示为1111的下一个数是2个1 表示为21
21的下一个数为1个2 1个1 表示为1211
以此类推,n表示第n个数。
题解
一个个迭代计算,迭代n次class Solution {
public:
string tran(string cur)
{
string res;
char key=cur[0];
int times=1;
for(int i=1;i!=cur.size();i++)
{
if(cur[i]==key)
times++;
else
{
char stimes=times+'0';
res=res+stimes+key;
key=cur[i];
times=1;
}
}
char stimes=times+'0';
res=res+stimes+key;
return res;
}
string countAndSay(int n) {
string cur="1";
while(--n)
cur=tran(cur);
return cur;
}
};
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