【Shawn-LeetCode】3. Longest Substring Without Repeating Characters-优快云博客

本文探讨了如何寻找字符串中最长的无重复字符子串，并提供了三种不同的解决方案，包括暴力破解、优化后的暴力破解及高效的滑动窗口算法。通过对比不同方法的时间复杂度，展示最优实践。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 asubstring, "pwke" is a subsequence and not a substring.

解决方案1：粗暴的直接遍历，O(n^4)

public static int lengthOfLongestSubstring(String s) {
		if(s.length()==0)return 0;
		int length=1;
		char[] a=s.toCharArray();
		int i=0,j=0,k=0,l=0;
		boolean flag=true;
		for(i=0;i<s.length()-1;i++){
			flag=true;
			A:
			for(j=i+1;j<s.length();j++){
				for(k=i;k<j;k++){
					for(l=k+1;l<=j;l++){
						if(a[k]==a[l]){
								flag=false;
								break A;
							}
					}
					
					
				}
				if(flag){
					int temp=j-i+1;
					System.out.println(temp+" "+j+" "+i);
					if(temp>length)length=temp;
				}
			}
		}
		return length;
    }

解决方案2：优化后，将长度作为变量进行循环，但还是没有本质上的区别，还是没有能够逃脱O（n^4）的厄运

public static int lengthOfLongestSubstring(String s) {
		if(s.length()==0)return 0;
		int length=1;
		int StringLength=s.length();//8
		int subLength=StringLength;
		int i=0;
		char[] a=s.toCharArray();
		A:
		for(;subLength>1;subLength--){
			
			for(i=0;i+subLength-1<StringLength;i++){
				if(checkRepeat(i,i+subLength-1,s)){
					//System.out.println(i+" "+(i+subLength-1));
					return subLength;
				}
			}
		}
		return length;
	}
	public static boolean checkRepeat(int start,int end,String s){
		char[] a=s.toCharArray();
		for(int i=start;i<end;i++){
			for(int j=i+1;j<=end;j++){
				if(a[i]==a[j])
					return false;
			}
		}
		return true;
	}

解决方案3：以上两种方案均在最后一组数据出错，最后一个数据为：

"abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!\"#$%&'()*+,-./:;<=>?@[\\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!\"#$%&'()*+,-./:;<=>?@[\\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!\"#$%&'()*+,-./:;<=>?@[\\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!\"#$%&'()*+,-./:;<=>?@[\\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!\"#$%&'()*+,-./:;<=>?@[\\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!\"#$%&'()*+,-./:;<=>?@[\\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!\"#$%&'()*+,-./:;<=>?@[\\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!\"#$%&'()*+,-./:;<=>?@[\\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!\"#$%&'()*+,-./:;<=>?@[\\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!\"#$%&'()*+,-./:;<=>?@[\\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!\"#$%&'()*+,-./:;<=>?@[\\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!\"#$%&'()*+,-./:;<=>?@[\\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!\"#$%&'()*+,-./:;<=>?@[\\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!\"#$%&'()*+,-./:;<=>?@[\\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!\"#$%&'()*+,-./:;<=>?@[\\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!\"#$%&'()*+,-./:;<=>?@[\\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!\"#$%&'()*+,-./:;<=>?@[\\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!\"#$%&'()*+,-./:;<=>?@[\\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!\"#$%&'()*+,-./:;<=>?@[\\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!\"#$%&'()*+,-./:;<=>?@[\\]^_`{|}~ abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!\"#$%&'()*+,-./:;<=>?@[\\]^_`{|}~ abcdefghij”

参考http://blog.youkuaiyun.com/waycaiqi/article/details/45936181，在此处也贴出他的代码，采用类似计算机网络中的窗口协议，通过hash表来快速的进行筛选，非常的高效，这组测试数据答案为95，时间较长但还是可以接受的范围。

    public class Solution {    
      public int lengthOfLongestSubstring(String s) {    
        if(s==null && s.length()==0)    
            return 0;    
              
        HashSet<Character> set = new HashSet<Character>();    
        int max = 0;    
        int walker = 0;    
        int runner = 0;    
        for(;runner<s.length();runner++)    
        {    
            if(set.contains(s.charAt(runner)))    
            {    
                max = (runner-walker)>max?(runner-walker):max;  
                  
                while(s.charAt(walker)!=s.charAt(runner))    
                {    
                    set.remove(s.charAt(walker));    
                    walker++;    
                }    
                walker++;    
            }    
            else    
            {    
                set.add(s.charAt(runner));    
            }    
        }    
        max = (runner-walker)>max?(runner-walker):max;    
        return max;    
       }    
    }

再贴一个学霸的代码，拿到授权了，就不贴地址了：

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        int c[256] = {0};
        int first = 0, last = 0;
        int len = 0;
        int maxlen = 0;
        for(int i = 0; i < s.size(); ++i){
            if(c[s[i]]){
                if(len > maxlen)
                    maxlen = len;
                last = c[s[i]];
                for(int j = first; j < last; ++j){
                    len--;
                    c[s[j]] = 0;
                }
                first = last;
            }
            len++;
            c[s[i]] = i + 1;
        }
        if(len > maxlen)
            maxlen = len;
        return maxlen;
    }
};