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Topic link and description
https://leetcode.cn/problems/largest-substring-between-two-equal-characters/
Given a string s, please return the length of the longest substring between two identical characters, excluding these two characters when calculating the length. Returns -1 if no such substring exists.
A substring is a contiguous sequence of characters within a string.
Example 1:
Input: s = "aa"
Output: 0
Explanation: The optimal substring is the empty substring between two 'a's.
Example 2:
Input: s = "abca"
Output: 2
Explanation: The optimal substring is "bc".
Example 3:
Input: s = "cbzxy"
Output: -1
Explanation: There is no character that appears twice in s, so return -1.
Example 4:
Input: s = "cabbac"
Output: 4
Explanation: The optimal substring is "abba", other non-optimal solutions include "bb" and "".
hint:
1 <= s.length <= 300
s contains only lowercase English letters
Keywords: sliding window double pointer hash table
Method 1: Sliding window double pointer
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the code
public int maxLengthBetweenEqualCharacters(String s) {
// 这是一道简单题，很容易想到双指针
// 从左右开始扫描 如果遇到相同的则是最优解
// 但是有个问题是，两边都可以动，那么需要使用滑动窗口来写
// 外层控制窗口大小，内层窗口左右滑动
for (int i = s.length() - 1; i > 0; i--) {
for (int head = 0, tail = i; tail < s.length(); head++, tail++) {
if (s.charAt(head) == s.charAt(tail)) {
return i - 1;
}
}
}
return -1;
}
Method 2: Hash Table
run screenshot
the code
class Solution {
public int maxLengthBetweenEqualCharacters(String s) {
// 还可以用hash表求得
// 因为只有26个小写字母
int[] letter = new int[26];
Arrays.fill(letter, s.length());
int res = -1;
for (int i = 0; i < s.length(); i++) {
// 转换成对应字母的int值
int u = s.charAt(i) - 'a';
// 对比字母最小的出现下标，赋值记录，所以前面填充就是为了记录下标
letter[u] = Math.min(letter[u], i);
// 拿到最小下标后和当前下标对比，得到当前字母的最大长度
// 然后与所有的最大取最大，即可得到最大
res = Math.max(res, i - letter[u] - 1);
}
return res;
}
}
end
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