Count and Say--数数
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Count and Say
题目 :
The count-and-say sequence is the sequence of integers beginning as follows:
1, 11, 21, 1211, 111221, ...
is read off as "one 1"
or .
is read off as "two 1s"
or .
is read off as "one 2
, then one 1"
or .
Given an integer n , generate the n th sequence.
Note: The sequence of integers will be represented as a string.
题 解 :
题目说的实在是太不明白了。。。
解释一下就是,输入n,那么我就打出第n行的字符串。
怎么确定第n行字符串呢?他的这个是有规律的。
n = 1时,打印一个1。
n = 2时,看n=1那一行,念:1个1,所以打印:11。
n = 3时,看n=2那一行,念:2个1,所以打印:21。
n = 4时,看n=3那一行,念:一个2一个1,所以打印:1211。
以此类推。(注意这里n是从1开始的)
所以构建当前行的字符串要依据上一行的字符串。 “小陷阱就是跑完循环之后记得把最后一个字符也加上,因为之前只是计数而已。”
public class Solution {
public String countAndSay(int n) {
String result = "1";
for (int outer = 1; outer < n; outer++) {
String previous = result;
result = "";
int count = 1;
char say = previous.charAt(0);
for (int i = 1; i < previous.length(); i++) {
if (previous.charAt(i) != say) {
result = result + count + say;
count = 1;
say = previous.charAt(i);
} else count++;
}
result = result + count + say;
}
return result;
}
}
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