leetcode-38 count and say

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题目的意思有点迷惑，正确的理解如下：

At the beginning, I got confusions about what is the nth sequence. Well, my solution is accepted now, so I'm going to give some examples of nth sequence here. The following are sequence from n=1 to n=10:

 1.     12.     113.     214.     12115.     111221 6.     3122117.     131122218.     11132132119.     3113121113122110.   13211311123113112211

From the examples you can see, the (i+1)th sequence is the "count and say" of the ith sequence!

从discuss中发现了一种很好的算法：

char *countAndSay(int n) {int i,j,k;char tmp[10240];char res[10240] = "1";for(i = 2; i <= n; i++){int count = 1;char digit = res[0];k = 0;for(j = 1; j < strlen(res); j++){if(res[j] == digit)count++;else{tmp[k++] = count + '0';tmp[k++] = digit;digit = res[j];count = 1;}}tmp[k++] = count + '0';tmp[k++] = digit;strncpy(res,tmp,k);}return res;
}

注意产生的串会很长的，所以数组tmp和res的长度要足够用（1024不够，如果使用1024，则一直runtime error）

char str[1024] = "1";

则strlen（str）的值为1；

sizeof（str）的值为1024；

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