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.
题意:统计上一个字符串中连续相同的字符的长度,求第n个思路:单纯的模拟
class Solution {
public:
string countAndSay(int n) {
string ans = "1";
for (int i = 1; i < n; i++) {
int count = 1;
string tmp = "";
for (int j = 1; j < ans.size(); j++) {
if (ans[j] == ans[j-1]) {
count++;
} else {
tmp = tmp + (char)(count + '0') + ans[j-1];
count = 1;
}
}
ans = tmp + (char)(count + '0') + ans[ans.size()-1];
}
return ans;
}
};
原文:http://blog.csdn.net/u011345136/article/details/44037151