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0984_string_without_aaa_or_bbb.java
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112 lines (101 loc) · 2.99 KB
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/*
* https://leetcode-cn.com/problems/string-without-aaa-or-bbb/
给定两个整数 A 和 B，返回任意字符串 S，要求满足：
1. S 的长度为 A + B，且正好包含 A 个 'a' 字母与 B 个 'b' 字母；
2. 子串 'aaa' 没有出现在 S 中；
3. 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。
----------------------------------------------------------------------------------------------------
Given two integers A and B, return any string S such that:
1. S has length A + B and contains exactly A 'a' letters, and exactly B 'b' letters;
2. The substring 'aaa' does not occur in S;
3. 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"
75AF
Note:
1. 0 <= A <= 100
2. 0 <= B <= 100
3. It is guaranteed such an S exists for the given A and B.
*/
class MySolution {
public String strWithout3a3b(int A, int B) {
StringBuilder sb = new StringBuilder();
int len = A + B;
int counterA = 0, counterB = 0;
for (int i = 0; i < len; ++i) {
if (counterA != 2 && counterB != 2) {
if (A > B) {
sb.append('a');
counterA++;
A--;
} else {
sb.append('b');
counterB++;
B--;
}
} else {
if (counterA == 2) {
sb.append('b');
counterA = 0;
counterB++;
B--;
} else {
sb.append('a');
counterB = 0;
counterA++;
A--;
}
}
}
return sb.toString();
}
}
class Solution1 {
public String strWithout3a3b(int A, int B) {
StringBuilder stringBuilder = new StringBuilder();
while (A > 0 && B > 0) {
if (A > B) {
stringBuilder.append("aab");
A -= 2;
B -= 1;
} else if (A == B) {
for (int i = 0; i < A; i++) {
stringBuilder.append("ab");
}
A = 0;
B = 0;
} else {
stringBuilder.append("bba");
A -= 1;
B -= 2;
}
}
if (A == 0) {
for (int i = 0; i < B; i++) {
stringBuilder.append("b");
}
} else {
for (int i = 0; i < A; i++) {
stringBuilder.append("a");
}
}
return stringBuilder.toString();
}
}