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：有一个整数序列 1,11,21,1211,111221,...
它是这样产生的：
1读作 ‘one 1’，即 11
11读作 ‘two 1s‘，即 21
21读作 ‘one 2 one 1‘，即1211
...
给定整数 n，返回该序列的第 n 个数
思路：模拟
复杂度：时间O(n)，空间O(1)
//一般写法
string countAndSay(int n){
string last = "1";
while(--n){
int count = 1;
char c = last[0];
string cur;
for(int i = 1; i <= last.size(); ++i){
if(i < last.size() && last[i] == c) count++;
else{
cur += count + '0';
cur += c;
if(i < last.size()){
count = 1;
c = last[i];
}
}
}
last = cur;
}
return last;
}
//使用stl
string countAndSay(int n){
string past = "1";
while(--n){
string cur;
for(auto iter = past.begin(); iter != past.end();){
auto iter2 = find_if(iter, past.end(), bind1st(not_equal_to<char>(), *iter));
cur += distance(iter, iter2) + '0';
cur += *iter;
iter = iter2;
}
past = cur;
}
return past;
}原文：http://blog.csdn.net/zhengsenlie/article/details/39121719