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.
我自己的代码：
public class Solution {
public int lengthOfLongestSubstring(String s) {
int slenmax=0;
for(int i=0;i<s.length();i++)
{
int j=i+1;
boolean flag=true;
while(j<s.length()&&flag)
{
for(int k=i;k<j;k++)
{
if(s.charAt(j)==s.charAt(k))
{
flag=false;
j--;
}
}
j++;
}
int slen=j-i;
if(slen>slenmax)
slenmax=slen;
}
return slenmax;
}
}
运行时可以通过，但是在提交时出现超时（思路比较简单，因而时间复杂度相应会比较高）
clean Code:
public class Solution {
public int lengthOfLongestSubstring(String s) {
boolean[] exist = new boolean[256]; //用表格处理，查找时会很快
int i = 0, maxLen = 0;
for (int j = 0; j < s.length(); j++) {
while (exist[s.charAt(j)]) { //这个while是精髓
exist[s.charAt(i)] = false;
i++;
}
exist[s.charAt(j)] = true;
maxLen = Math.max(j - i + 1, maxLen);
}
return maxLen;
}
}
注：借用了表的形式，把字符串中的字母依次存入表中并进行相应标记（如 exist[s.charAt(i)] = false;）。解决思路是：设定两个指针（i, j），都放在左头开始，从左到右，遇到有两个相同字母的情况，则计算两相同字母间距，更新最大间距，并且把i 也放在j 处，再进行计算最大间距。
Leetcode 详解（Substing without repeats character）
原文：http://www.cnblogs.com/zehua-shu/p/5424125.html