38.Count-and-Say

38. Count and Say

题目地址

https://leetcode.com/problems/count-and-say/

题目描述

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".

代码

Approach #1

class Solution {
  public String countAndSay(int n) {
        LinkedList<Integer> prevSeq = new LinkedList<Integer>();
    prevSeq.add(1);
    // Using -1 as the delimiter
    prevSeq.add(-1);

    List<Integer> finalSeq = nextSequence(n, prevSeq);
    StringBuilder seqStr = new StringBuilder();
    for (Integer digit: finalSeq) {
      seqStr.append(String.valueOf(digit));
    }
    return seqStr.toString();
  }

  private LinkedList<Integer> nextSequence(int n, LinkedList<Integer> prevSeq) {
    if (n <= 1) {
      // remove the delimiter before return
      prevSeq.pollLast();
      return prevSeq;
    }

    LinkedList<Integer> nextSeq = new LinkedList();
    Integer prevDigit = null;
    Integer digitCnt = 0;
    for (Integer digit: prevSeq) {
      if (prevDigit == null) {
        prevDigit = digit;
        digitCnt += 1;
      } else if (digit == prevDigit) {
        digitCnt += 1;
      } else {
             // digit != prevDigit: the last digit == -1 will be ignored
        nextSeq.add(digitCnt);
        nextSeq.add(prevDigit);
        // reset
        prevDigit = digit;
        digitCnt = 1;
      }
    }

    nextSeq.add(-1);
    return nextSequence(n - 1, nextSeq);
  }
}

Approach #2 Two loop

public class Solution {
    public String countAndSay(int n) {
        if(n <= 0) return "-1";
        String result = "1";

        for(int i = 1; i < n; i ++) {
            result = build(result);
        }
        return result;
    }

  private String build(String result) {
    StringBuilder builder = new StringBuilder();
    int p = 0;
    while (p < result.length()) {
        char val = result.charAt(p);
        int count = 0;
        while (p < result.length() && 
          result.charAt(p) == val){
            p ++;
            count ++;
        }
        builder.append(String.valueOf(count));
        builder.append(val);
    }
    return builder.toString();
  }
}