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Leetcode 3 Longest Substring Without Repeating Characters 题解分析

又做了个题,看记录是以前用 C++写过的,现在捋一捋思路,用 Java 再写了一下,思路还是比较清晰的,但是边界细节处理得比较差

简要介绍

Given a string s, find the length of the longest substring without repeating characters.

样例

Example 1:

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Input: s = "abcabcbb"
Output: 3
Explanation: The answer is "abc", with the length of 3.

Example 2:

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Input: s = "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.

Example 3:

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Input: s = "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3.
Notice that the answer must be a substring, "pwke" is a subsequence and not a substring.

Example 4:

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Input: s = ""
Output: 0

就是一个最长不重复的字符串长度,因为也是中等难度的题,不太需要特别复杂的思考,最基本的就是O(N*N)两重循环,不过显然不太好,万一超时间,还有一种就是线性复杂度的了,这个就是需要搞定一个思路,比如字符串时 abcdefgaqwrty,比如遍历到第二个a的时候其实不用再从头去遍历了,只要把前面那个a给排除掉,继续往下算就好了

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class Solution {
Map<String, Integer> counter = new HashMap<>();
public int lengthOfLongestSubstring(String s) {
int length = s.length();
// 当前的长度
int subStringLength = 0;
// 最长的长度
int maxSubStringLength = 0;
// 考虑到重复的位置已经被跳过的情况,即已经在当前长度的字符串范围之前的重复字符不需要回溯
int lastDuplicatePos = -1;
for (int i = 0; i < length; i++) {
// 使用 map 存储字符和上一次出现的位置,如果存在并且大于上一次重复位置
if (counter.get(String.valueOf(s.charAt(i))) != null && counter.get(String.valueOf(s.charAt(i))) > lastDuplicatePos) {
// 记录重复位置
lastDuplicatePos = counter.get(String.valueOf(s.charAt(i)));
// 重置不重复子串的长度,减去重复起点
subStringLength = i - counter.get(String.valueOf(s.charAt(i))) - 1;
// 替换当前位置
counter.replace(String.valueOf(s.charAt(i)), i);
} else {
// 如果不存在就直接 put
counter.put(String.valueOf(s.charAt(i)), i);
}
// 长度累加
subStringLength++;
if (subStringLength > maxSubStringLength) {
// 简单替换
maxSubStringLength = subStringLength;
}
}
return maxSubStringLength;
}
}

注释应该写的比较清楚了。

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