[Leetcode] Count and Say 数个数

Count Consecutive Digits in Integer

Count consecutive digits and say it. For example, return 132341 if input is 1112224. There are three 1s, three 2s and one 4.

反转字符法

复杂度

时间 O(N) 空间 O(1)

思路

因为数字不好从前向后遍历每一位（要先统计一共有多少位，比较麻烦），所以我们直接从后向前计数，最后把结果倒置就行了。

注意

退出循环后还要额外执行一次append，将最后的连续数字补齐

代码

public String countAndSay(int n) {
    if(n <= 0) return "";
    int last = n % 10, cnt = 1;
    n = n / 10;
    StringBuilder sb = new StringBuilder();
    while(n > 0){
        int digit = n % 10;
        if(digit == last){
            cnt++;
        } else {
            sb.append(cnt);
            sb.append(last);
            cnt = 1;
            last = digit;
        }
        n = n / 10;
    }
    sb.append(cnt);
    sb.append(last);
    sb.reverse();
    return sb.toString();
}

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.

递归法

复杂度

时间 O(N) 空间 O(N) 递归栈空间

思路

因为第n个数count and say的结果是基于第n-1个数的，我们可以用递归解决这个问题。

代码

public class Solution {
    public String countAndSay(int n) {
        if(n == 0){
            return "";
        }
        if(n == 1){
            return "1";
        }
        String s = countAndSay(n-1);
        char last = s.charAt(0);
        int cnt = 1;
        StringBuilder sb = new StringBuilder();
        for(int i = 1; i < s.length(); i++){
            if(s.charAt(i)==last){
                cnt++;
            } else {
                sb.append(cnt);
                sb.append(last);
                cnt = 1;
                last = s.charAt(i);
            }
        }
        sb.append(cnt);
        sb.append(last);
        return sb.toString();
    }
}