[leetcode] Count and Say

Count and Say

The count-and-say sequence is the sequence of integers beginning as follows:

1, 11, 21, 1211, 111221, ...

1

is read off as

"one 1"

or

11

.

11

is read off as

"two 1s"

or

21

.

21

is 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.

思路：

举例说明怎么做，求 s=1211 的下一个，用res记录。从下标1开始遍历这个字符串，下标0作为比较字符。将s[1]与s[0]比较，不等，于是s[0]只重复了一次，因此就是1个s[0]，此时res=11。再将s[2]与s[1]比较，还是不等，res=1112。s[3]与s[2]比较，相等，即s[2]重复了两次，也就是2个s[2]，此时res=111221。有一个需要注意的地方，遍历s的时候最后添加的是倒数第二个不重复的字符，最后的那些重复的字符只是与前面进行比较并没有添加到res中。因此在遍历完成后，还需要添加最后那些重复的字符。

题解：

class Solution {
public:
string fun(string s) {
string str;
int count = 1;
char pre = s[0];
char tmp;
for(int i=1;i<s.size();i++) {
if(s[i]==pre)
count++;
else {
tmp = count+'0';
str = str+tmp+pre;
count = 1;
pre = s[i];
}
}
tmp = count+'0';
str = str+tmp+pre;
return str;
}
string countAndSay(int n) {
string res = "1";
for(int i=1;i<n;i++)
res = fun(res);
return res;
}
};

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