[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.

题意： 读数字，开始是1，然后是两一个一(1 1),然后是两个一(2 1)，然后是一个二一个一(1 2 1 1).求第n次的结果，存成字符串返回.

解法： 嘛,其实感觉这题没意思，因为数字太大的话，时间会很长.数据小的话，就没意思了，因为数字的连续数量不会超过9个,总之，直接模拟就好了，用两个串来记当前串和下一串，读当前串结果存在下一串，然后读下一串，读的过程就是两个for循环，第一重是枚举第i个数，然后跳到下一个与i不同的数，第二重就是一直循环把相同的数跑完就好了，然后用j-i得出连续的数有多少个.复杂度：O(n*m)m在递增，一次比一次长.数据大概应该是只有1到18吧.18组测试数据.

class Solution {
public:
string countAndSay(int n) {
string akResult[2];
int iCurIdx = 0;
for (akResult[iCurIdx].push_back('1'); --n; iCurIdx ^= 1) {
akResult[iCurIdx ^ 1].clear();
for (int i = 0, j = 0; i < (int)akResult[iCurIdx].size(); i = j) {
for (j = i; j < (int)akResult[iCurIdx].size() && akResult[iCurIdx][j] == akResult[iCurIdx][i]; ++j);
akResult[iCurIdx ^ 1].push_back((j - i) + '0');
akResult[iCurIdx ^ 1].push_back(akResult[iCurIdx][i]);
}
}
return akResult[iCurIdx];
}
};

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