3 Longest Substring Without Repeating Characters 无重复字符的最长子串
Description:
Given a string, find the length of the longest substring without repeating characters.
Example:
Example 1:
Input: "abcabcbb"
Output: 3
Explanation: The answer is "abc", with the length of 3.
Example 2:
Input: "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.
Example 3:
Input: "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3.
Note that the answer must be a substring, "pwke" is a subsequence and not a substring.
题目描述:
给定一个字符串，请你找出其中不含有重复字符的 最长子串 的长度。
示例 :
示例 1:
输入: "abcabcbb"
输出: 3
解释: 因为无重复字符的最长子串是 "abc"，所以其长度为 3。
示例 2:
输入: "bbbbb"
输出: 1
解释: 因为无重复字符的最长子串是 "b"，所以其长度为 1。
示例 3:
输入: "pwwkew"
输出: 3
解释: 因为无重复字符的最长子串是 "wke"，所以其长度为 3。
请注意，你的答案必须是 子串 的长度，"pwke" 是一个子序列，不是子串。
思路:
滑动窗口法, 设置左指针 l = 0, 右指针 r = -1, 移动右指针保证右指针不移出字符串 s, 用一个长度为128(长度为26也可)记录出现过的字符, 如果这个字符串出现过(数组记录不为 0), 则移动左指针减少滑动窗口的长度, 否则移动右指针增加滑动窗口长度, 记录最大的滑动窗口长度即可
时间复杂度O(n), 空间复杂度O(1), ASCII码最多128个需要记录
代码:
C++:
class Solution
{
public:
int lengthOfLongestSubstring(string s)
{
int l = 0, r = -1, result = 0, count[128]{0};
while (l < s.size())
{
if (r + 1 < s.size() and !count[s[r + 1]]) ++count[s[++r]];
else --count[s[l++]];
result = max(result, r - l + 1);
}
return result;
}
};
Java:
class Solution {
public int lengthOfLongestSubstring(String s) {
int l = 0, r = -1, result = 0, count[] = new int[128];
while (l < s.length()) {
if (r + 1 < s.length() && count[s.charAt(r + 1)] == 0) ++count[s.charAt(++r)];
else --count[s.charAt(l++)];
result = Math.max(result, r - l + 1);
}
return result;
}
}
Python:
class Solution:
def lengthOfLongestSubstring(self, s: str) -> int:
l, r, result, count = 0, -1, 0, [False] * 128
while l < len(s):
if r + 1 < len(s) and not count[ord(s[r + 1])]:
r += 1
count[ord(s[r])] = True
else:
count[ord(s[l])] = False
l += 1
result = max(result, r - l + 1)
return result