leetcode刷题笔记-Longest Substring Without Repeating Characters

问题描述

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

Examples:

Given “abcabcbb”, the answer is “abc”, which the length is 3.

Given “bbbbb”, the answer is “b”, with the length of 1.

Given “pwwkew”, 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.

最大不重复子串问题

解答思路

直接使用两层循环，内层循环判断当前子串的后一个字符是否已在当前子串中

public int lengthOfLongestSubstring(String s) {
int max;
if (s.length() == 0) {
max = 0;
} else {
max = 1;
}
int length = 0;
String sub;
for (int i = 0; i < s.length() - 1; i++) {
sub = s.substring(i, i + 1);
for (int j = i + 1; j < s.length(); j++) {
if (sub.indexOf(s.charAt(j)) != -1) {
break;
} else {
sub = s.substring(i, j + 1);
}
}
length = sub.length();
max = max > length ? max : length;
}
return max;
}

改进的算法：使用滑动窗口

public class Solution {
public int lengthOfLongestSubstring(String s) {
int n = s.length();
Set<Character> set = new HashSet<>();
int ans = 0, i = 0, j = 0;
while (i < n && j < n) {
if (!set.contains(s.charAt(j))){
set.add(s.charAt(j++));
ans = Math.max(ans, j - i);
}
else {
set.remove(s.charAt(i++));
}
}
return ans;
}
}

i，j分别为滑动窗口的起始位置和结束位置，判断第j个字符是否在当前子串中，如果不在，则将第j个字符放入set中，然后j++，判断下一个字符，如果第j个字符在set中，则从子串中移除第i个字符，继续判断第j个字符是否在子串中