3. Longest Substring Without Repeating Characters 不含重复字母的最长子串
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2024-02-26 14:27:16
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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.
给定一个字符串,找出其中不含重复字母的最长子串(注意不是子序列),返回它的长度。
思路
字母查重:构建一bool列表,存储当前子串中是否含有某字母。
设置两指针i和j,分别表示子串头尾,j向后扫描,每次进行查重,保存当前子串长度,遇重复字母则i移动到此字母第一次出现的下一个位置,并将移除子串的字母进行查重表更新。
class Solution {
public:
int lengthOfLongestSubstring(string s) {
int i=0,j=0;
int n=s.size();
int maxlen=0;
bool exist[256]={false};
while(j<n){
if(exist[s[j]]){
maxlen=max(maxlen,j-i);
while(s[i]!=s[j]){
exist[s[i++]]=false;
}
++i;
++j;
}
else{
exist[s[j++]]=true;
}
}
maxlen=max(maxlen,n-i);
return maxlen;
}
};
也可使用HashMap存储字符出现的位置,若存在则取同一个字符两个位置的距离,若不存在则记录字符的位置。
public class Solution {
public int lengthOfLongestSubstring(String s) {
if (s.length()==0) return 0;
HashMap<Character, Integer> map = new HashMap<Character, Integer>();
int res=0;
for (int i=0, previous=0; i<s.length(); ++i){
if (map.containsKey(s.charAt(i))){
previous = Math.max(previous,map.get(s.charAt(i))+1);
}
map.put(s.charAt(i),i);
res = Math.max(res,i-previous+1);
}
return res;
}
}
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