题目描述:
Given a string, find the length of the longest substring without repeating characters.
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.
中文理解:给定一个字符串,求出字符串中不含重复字符的子序列的最大值。
解题思路:
解答一:
使用二重循环,外循环假设为初始的字符的下标,内循环从外循环开始,获取最大不重复子串的长度,若发现重复,跳出,与全局最大值比较,获得最大值。
代码(java):
class Solution {
public int lengthOfLongestSubstring(String s) {
if(s.length()<=1)return s.length();
int max=-1;
for(int i=0;i<s.length();i++){
HashSet<Character> set=new HashSet<Character>();
int unique=0;
for(int j=i;j<s.length();j++){
if(set.contains(s.charAt(j))){
break;
}
else{
unique++;
set.add(s.charAt(j));
}
}
max=Math.max(max,unique);
}
return max;
}
}
解法二:
使用两个指针来进行模拟,i,j两个指针,其中i指针在前,j指针在后,使用HashMap存储字符,下标,如果字符在子串中重复出现,则更新j值,同时每次修改max值来进行获取全局最大值。
代码(java):
class Solution {
public int lengthOfLongestSubstring(String s) {
if(s.length()<=1)return s.length();
HashMap<Character,Integer> map=new HashMap<Character,Integer>();
int max=-1;
for(int i=0,j=0;i<s.length();i++){
if(map.keySet().contains(s.charAt(i))){
j=Math.max(j,map.get(s.charAt(i))+1);
}
map.put(s.charAt(i),i);
max=Math.max(max,i-j+1);
}
return max;
}
}