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.
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以上含义描述不清,题目的实际意思是:
n = 0: 1
n = 1: 11 (前一个是1个1)
n = 2: 21 (前一个是2个1)
n = 3: 1211 (前一个是1个2,1个1)
n = 4: 111221 (前一个是1个1,1个2,2个1)
....
所以是一个递推关系,按规则模拟即可。
1 class Solution { 2 public: 3 string countAndSay(int n) { 4 string result("1"); 5 for (int i = 1; i < n; ++i) { 6 stringstream ss; 7 int j = 0, N = result.size(); 8 while (j < N) { 9 int c = 1; 10 while (j + 1 < N && result[j] == result[j + 1]) { 11 ++c; 12 ++j; 13 } 14 ss << c << result[j]; 15 ++j; 16 } 17 ss >> result; 18 } 19 return result; 20 } 21 };
【Leetcode】Count and Say,布布扣,bubuko.com
原文地址:http://www.cnblogs.com/dengeven/p/3766000.html