Leetcode 481. Magical String 魔幻字符串解题报告

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本文介绍了LeetCode中481题——魔幻字符串的解题思路。题目要求理解一个由1和2组成的字符串，其特性是连续的1或2分组后的计数字符串与其自身相同。虽然作者没有亲自解答，但分享了AC（Accepted）解决方案。

1 解题思想

这道题的就是，有一个只包含1和2的字符串，他有一个神奇的特性：
*如果将字符串按照1和2分组后，统计每组1或2出现的次数所拼接成的字符串，和这个字符串本身一样

我没做这道题。。感觉太烦，直接贴的别人的答案了，有兴趣我再来自己写

2 原题

A magical string S consists of only '1' and '2' and obeys the following rules:

The string S is magical because concatenating the number of contiguous occurrences of characters '1' and '2' generates the string S itself.

The first few elements of string S is the following: S = "1221121221221121122……"

If we group the consecutive '1's and '2's in S, it will be:

1 22 11 2 1 22 1 22 11 2 11 22 ......

and the occurrences of '1's or '2's in each group are:

1 2 2 1 1 2 1 2 2 1 2 2 ......

You can see that the occurrence sequence above is the S itself.

Given an integer N as input, return the number of '1's in the first N number in the magical string S.

Note: N will not exceed 100,000.

Example 1:
Input: 6
Output: 3
Explanation: The first 6 elements of magical string S is "12211" and it contains three 1's, so return 3.

3 AC解

//此题来自Discuss的解
public class Solution {
    public int magicalString(int n) {
        StringBuilder magic = new StringBuilder("1221121221221121122");
        int pt1 = 12, pt2 = magic.length(), count = 0; 
        while(magic.length() < n){
            if(magic.charAt(pt1) == '1'){
                if(magic.charAt(pt2-1) == '1') magic.append(2);
                else magic.append(1);
                pt2++;
            }else{ //==2
                if(magic.charAt(pt2-1) == '1') magic.append(22);
                else magic.append(11);
                pt2+=2;
            }
            pt1++;
        }
        for(int i=0;i<n;i++)
            if(magic.charAt(i)=='1') count++;
        return count;

    }
}