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.

給定兩個整數 A 和 B，返回任意字符串 S，要求滿足：

• S 的長度為 A + B，且正好包含 A 個 ‘a‘ 字母與 B 個 ‘b‘ 字母；
• 子串 ‘aaa‘ 沒有出現在 S 中；
• 子串 ‘bbb‘ 沒有出現在 S 中。

示例 1：

輸入：A = 1, B = 2
輸出："abb"
解釋："abb", "bab" 和 "bba" 都是正確答案。

示例 2：

輸入：A = 4, B = 1
輸出："aabaa" 

提示：

1. 0 <= A <= 100
2. 0 <= B <= 100
3. 對於給定的 A 和 B，保證存在滿足要求的 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 }

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