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38. Count and Say
看懂了这道题，相声就已经入门了……
Matrix67 的解释 Conway常数是怎么得来的？ | Matrix67: The Aha Moments 还是非常浅显易懂的！
The count-and-say sequence is the sequence of integers with the first five terms as following:
1. 1 2. 11 3. 21 4. 1211 5. 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 where 1 ≤ n ≤ 30, generate the nth term of the count-and-say sequence. You can do so recursively, in other words from the previous member read off the digits, counting the number of digits in groups of the same digit.
Note: Each term of the sequence of integers will be represented as a string.
Example 1:
Input: 1 Output: "1" Explanation: This is the base case.
Example 2:
Input: 4 Output: "1211" Explanation: For n = 3 the term was "21" in which we have two groups "2" and "1", "2" can be read as "12" which means frequency = 1 and value = 2, the same way "1" is read as "11", so the answer is the concatenation of "12" and "11" which is "1211".
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/**
* Runtime: 3 ms, faster than 40.24% of Java online submissions for Count and Say.
*
* Memory Usage: 40.9 MB, less than 5.26% of Java online submissions for Count and Say.
*/
public String countAndSay(int n) {
String result = "1";
if (n == 1) {
return result;
}
StringBuilder builder;
for (int i = 1; i < n; i++) {
builder = new StringBuilder();
char[] chars = result.toCharArray();
Character prefix = null;
int count = 0;
for (int j = 0; j < chars.length; j++) {
char current = chars[j];
if (Objects.isNull(prefix)) {
prefix = current;
++count;
} else {
if (Objects.equals(prefix, current)) {
++count;
} else {
builder.append(count).append(prefix);
prefix = current;
count = 1;
}
}
if (j == chars.length - 1) {
builder.append(count).append(prefix);
}
}
result = builder.toString();
}
return result;
}