1156 Swap For Longest Repeated Character Substring 单字符重复子串的最大长度
Description:
You are given a string text. You can swap two of the characters in the text.
Return the length of the longest substring with repeated characters.
Example:
Example 1:
Input: text = "ababa"
Output: 3
Explanation: We can swap the first 'b' with the last 'a', or the last 'b' with the first 'a'. Then, the longest repeated character substring is "aaa" with length 3.
Example 2:
Input: text = "aaabaaa"
Output: 6
Explanation: Swap 'b' with the last 'a' (or the first 'a'), and we get longest repeated character substring "aaaaaa" with length 6.
Example 3:
Input: text = "aaaaa"
Output: 5
Explanation: No need to swap, longest repeated character substring is "aaaaa" with length is 5.
Constraints:
1 <= text.length <= 2 * 10^4
text consist of lowercase English characters only.
题目描述:
如果字符串中的所有字符都相同，那么这个字符串是单字符重复的字符串。
给你一个字符串 text，你只能交换其中两个字符一次或者什么都不做，然后得到一些单字符重复的子串。返回其中最长的子串的长度。
示例 :
示例 1：
输入：text = "ababa"
输出：3
示例 2：
输入：text = "aaabaaa"
输出：6
示例 3：
输入：text = "aaabbaaa"
输出：4
示例 4：
输入：text = "aaaaa"
输出：5
示例 5：
输入：text = "abcdef"
输出：1
提示:
1 <= text.length <= 20000
text 仅由小写英文字母组成。
思路:
模拟
只会出现 2 种情况
一段连续的加上其他地方的, 如 ...aaaa...a..., 可以移动之后形成 ...aaaaa..., 总数为连续最长的加 1
一段连续的中间只有一个其他字符, 如 ...aaabaaa...a..., 可以将 b 和 a 交换形成 ...aaaaaaa...b..., 总数为两端的总和加 1
先记录所有字符出现的次数
然后记录相同字符出现的长度
一边遍历一边更新结果
时间复杂度为 O(n), 空间复杂度为 O(n)
代码:
C++:
class Solution
{
public:
int maxRepOpt1(string text)
{
int n = text.size(), result = 0, idx = -1, nums[n];
unordered_map<int, int> pos;
for (const auto c : text) ++pos[c];
for (int i = 0; i < n; i++)
{
int j = i;
while (j < n and text[j] == text[i]) ++j;
nums[++idx] = j - i;
result = (idx > 1 and nums[idx - 1] == 1 and i > 1 and text[i - 2] == text[i]) ? max(result, min(nums[idx] + nums[idx - 2] + 1, pos[text[i]])) : max(result, min(nums[idx] + 1, pos[text[i]]));
i = j - 1;
}
return result;
}
};
Java:
class Solution {
public int maxRepOpt1(String text) {
int n = text.length(), result = 0, idx = -1, nums[] = new int[n];
Map<Character, Integer> pos = new HashMap<>();
for (char c : text.toCharArray()) pos.put(c, pos.getOrDefault(c, 0) + 1);
for (int i = 0; i < n; i++) {
int j = i;
while (j < n && text.charAt(j) == text.charAt(i)) ++j;
nums[++idx] = j - i;
result = (idx > 1 && nums[idx - 1] == 1 && i > 1 && text.charAt(i - 2) == text.charAt(i)) ? Math.max(result, Math.min(nums[idx] + nums[idx - 2] + 1, pos.get(text.charAt(i)))) : Math.max(result, Math.min(nums[idx] + 1, pos.get(text.charAt(i))));
i = j - 1;
}
return result;
}
}
Python:
class Solution:
def maxRepOpt1(self, text: str) -> int:
result, idx, nums, i, c = 0, -1, [0] * (n := len(text)), 0, Counter(text)
while i < n:
j = i
while j < n and text[j] == text[i]:
j += 1
idx += 1
nums[idx] = j - i
result = max(result, min(nums[idx] + nums[idx - 2] + 1, c[text[i]])) if idx > 1 and nums[idx - 1] == 1 and i > 1 and text[i - 2] == text[i] else max(result, min(nums[idx] + 1, c[text[i]]))
i = j
return result
