[Swift Weekly Contest 121]LeetCode984. 不含 AAA 或 BBB 的字符串 | String Without AAA or BBB

Given two integers A and B,return any string S such that:

• S has length A + B and contains exactly A ‘a‘ letters,and exactly B ‘b‘ letters;
• The substring ‘aaa‘ does not occur in S;
• The substring ‘bbb‘ does not occur in S. 

Example 1:

Input: A = 1,B = 2 Output: "abb" Explanation: "abb","bab" and "bba" are all correct answers. 

Example 2:

Input: A = 4,B = 1 Output: "aabaa" 

Note:

1. 0 <= A <= 100
2. 0 <= B <= 100
3. It is guaranteed such an S exists for the given A and B.

给定两个整数 A 和 B，返回任意字符串 S，要求满足：

• S 的长度为 A + B，且正好包含 A 个 ‘a‘ 字母与 B 个 ‘b‘ 字母；
• 子串 ‘aaa‘ 没有出现在 S 中；
• 子串 ‘bbb‘ 没有出现在 S 中。 

示例 1：

输入：A = 1,B = 2
输出："abb"
解释："abb","bab" 和 "bba" 都是正确答案。

示例 2：

输入：A = 4,B = 1
输出："aabaa" 

提示：

1. 0 <= A <= 100
2. 0 <= B <= 100
3. 对于给定的 A 和 B，保证存在满足要求的 S。

12ms

 1 class Solution {
 2     func strWithout3a3b(_ A: Int,_ B: Int) -> String {
 3         var A = A
 4         var B = B
 5         var ret:[Character] = [Character](repeating:" ",count:A+B)
 6         for i in 0..<ret.count
 7         {
 8             if i >= 2 && ret[i-1] == ret[i-2]
 9             {
10                 if ret[i-1] == "a"
11                 {
12                     ret[i] = "b"
13                     B -= 1
14                 }
15                 else
16                 {
17                     ret[i] = "a"
18                     A -= 1
19                 }
20             }
21             else
22             {
23                 if A > B
24                 {
25                     ret[i] = "a"
26                     A -= 1
27                 }
28                 else
29                 {
30                     ret[i] = "b"
31                     B -= 1
32                 }
33             }            
34         }
35         return String(ret.reversed())
36     }
37 }