[LeetCode] Magical String 神奇字符串
2017-01-15 03:50
288 查看
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:
这道题介绍了一种神奇字符串,只由1和2组成,通过计数1组和2组的个数,又能生成相同的字符串。而让我们求前n个数字中1的个数说白了其实就是让我们按规律生成这个神奇字符串,只有生成了字符串的前n个字符,才能统计出1的个数。其实这道题的难点就是在于找到规律来生成字符串,这里我们就直接说规律了,因为博主也没有自己找到,都是看了网上大神们的解法。根据第三个数字2开始往后生成数字,此时生成两个1,然后根据第四个数字1,生成一个2,再根据第五个数字1,生成一个1,以此类推,生成的数字1或2可能通过异或3来交替生成,在生成的过程中同时统计1的个数即可,参见代码如下:
解法一:
下面这种解法的思路跟上面一样,但是写法上面大大的简洁了,感觉很叼!
解法二:
参考资料:
https://discuss.leetcode.com/topic/74637/short-c
https://discuss.leetcode.com/topic/74917/simple-java-solution-using-one-array-and-two-pointers
LeetCode All in One 题目讲解汇总(持续更新中...)
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.
这道题介绍了一种神奇字符串,只由1和2组成,通过计数1组和2组的个数,又能生成相同的字符串。而让我们求前n个数字中1的个数说白了其实就是让我们按规律生成这个神奇字符串,只有生成了字符串的前n个字符,才能统计出1的个数。其实这道题的难点就是在于找到规律来生成字符串,这里我们就直接说规律了,因为博主也没有自己找到,都是看了网上大神们的解法。根据第三个数字2开始往后生成数字,此时生成两个1,然后根据第四个数字1,生成一个2,再根据第五个数字1,生成一个1,以此类推,生成的数字1或2可能通过异或3来交替生成,在生成的过程中同时统计1的个数即可,参见代码如下:
解法一:
class Solution { public: int magicalString(int n) { if (n <= 0) return 0; if (n <= 3) return 1; int res = 1, head = 2, tail = 3, num = 1; vector<int> v{1, 2, 2}; while (tail < n) { for (int i = 0; i < v[head]; ++i) { v.push_back(num); if (num == 1 && tail < n) ++res; ++tail; } num ^= 3; ++head; } return res; } };
下面这种解法的思路跟上面一样,但是写法上面大大的简洁了,感觉很叼!
解法二:
class Solution { public: int magicalString(int n) { string s = "122"; int i = 2; while (s.size() < n) { s += string(s[i++] - '0', s.back() ^ 3); } return count(s.begin(), s.begin() + n, '1'); } };
参考资料:
https://discuss.leetcode.com/topic/74637/short-c
https://discuss.leetcode.com/topic/74917/simple-java-solution-using-one-array-and-two-pointers
LeetCode All in One 题目讲解汇总(持续更新中...)
相关文章推荐
- leetcode 481. Magical String 神奇字符串+有点绕的题
- 字符串反转,神奇的算法-读《编程珠玑(第二版)》
- Wonderful String Translation Function in Python / Python神奇的字符串变换函数
- Longest Palindromic Substring 在一个字符串里找最长回文子串@LeetCode
- 神奇的字符串处理函数——sscanf()
- LeetCode 之 string字符串一
- 【题解】【字符串】【Leetcode】Valid Palindrome
- Palindrome Partitioning 分割字符串为回文@LeetCode
- Multiply Strings 两个字符串代表数字相乘@LeetCode
- 两个字符串交叉得到的字符串 Interleaving String @LeetCode
- 字符串反转,神奇的算法-读《编程珠玑(第二版)》 推荐
- Implement strStr() (匹配字符串问题,哈希解法) 【leetcode】
- leetcode:Longest Palindromic Substring(求最大的回文字符串)
- Longest Common Prefix 字符串数组中查找最长相同前缀字符串@LeetCode
- leetcode:Multiply Strings(字符串的乘法)【面试算法题】
- 字符串常量的神奇使用
- Duff's Device 神奇的编程 将字符串拷贝性能提升8倍
- Submission Details (两个字符串的最小距离)【leetcode】
- ZigZag Conversion 把一个字符串转为zigzag的表示形式@LeetCode
- 神奇的字符串