【leetcode】984. String Without AAA or BBB

题目如下：

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 Aand B.

解题思路：由于aaa/bbb是不允许的，因此Output必定只能由a,aa,b,bb四种组合。假设这四种组合的出现次数分别是i，j，k，l。那么应该满足 i+2*j = A，k+2*l = B，abs((i+j) - (k+l)) <= 1 （两者的差必须小于等于1，否则可能会出现aaa/bbb的情况）。 接下来只要穷举，找出其中一个符合条件的组合即可。

代码如下：

class Solution(object):
    def strWithout3a3b(self, A, B):
        """
        :type A: int
        :type B: int
        :rtype: str
        """
        a_1 = A
        a_2 = A/2
        b_1 = B
        b_2 = B/2
        def calc(a_1,a_2,b_1,b_2):
            for i in range(a_1+1):
                for j in range(a_2+1):
                    if i + 2*j != A:
                        continue
                    for k in range(b_1+1):
                        for l in range(b_2+1):
                            if k + 2*l != B:
                                continue
                            if abs(i+j-k-l) <= 1:
                                return i,j,k,l
        i,j,k,l = calc(a_1,a_2,b_1,b_2)
        res = ''
        next = 'a' if (i+j) >= (k+l) else 'b'
        while i + j + k + l > 0:
            if next == 'a':
                if i > 0:
                    res += next
                    i -= 1
                else:
                    res += next*2
                    j -= 1
                next = 'b'
            else:
                if k > 0:
                    res += next
                    k -= 1
                else:
                    res += next*2
                    l -= 1
                next = 'a'
        return res