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481 Magical String

1. Question

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 stringSis the following:S= "1221121221221121122……"
If we group the consecutive '1's and '2's inS, 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 theSitself.
Given an integer N as input, return the number of '1's in the first N number in the magical stringS.
Note:N will not exceed 100,000.
Example 1:
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Input: 6
2
3
Output: 3
4
5
Explanation:
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The first 6 elements of magical string S is "12211" and it contains three 1's, so return 3.
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2. Implementation

(1) Two Pointers
思路: 这道题的关键在于找到形成magical string的规律,这里的规律是从"122"中的第3位开始,,该位置上是数字2,先形成2个1,然后到第四位,该位置上是数字1,形成1个2,然后第5位,该位置上是1, 形成1个1,以此类推
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class Solution {
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public int magicalString(int n) {
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if (n <= 0) {
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return 0;
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}
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if (n <= 3) {
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return 1;
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}
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int[] sequence = new int[n + 1];
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sequence[0] = 1;
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sequence[1] = 2;
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sequence[2] = 2;
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int head = 2, tail = 3;
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int res = 1;
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int num = 1;
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while (tail < n) {
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for (int i = 0; i < sequence[head]; i++) {
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sequence[tail] = num;
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if (num == 1 && tail < n) {
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++res;
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}
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++tail;
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}
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num ^= 3;
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++head;
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}
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return res;
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}
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}
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3. Time & Space Complexity

Two Pointers: 时间复杂度O(n), 空间复杂度O(n)