[leetcode]3. Longest Substring Without Repeating Characters

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.

本题的解题思路就是设置两个指针m，n和一个hash表记录字符的索引，两个同时从字符串的第一个字符出发，如果hash表中没有n所在位置的字符信息，n不断向后移动。否则如果hash表中有n的字符的信息，处理m：

m用来记录每次计算当前局部最大子串长度时子串的开头字符的位置。如果m的值小于等于n所在字符出现的上一个索引，则m=上一个索引+1。

Java代码

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        if(s==null){
            return 0;
        }
        if(s.length()==0||s.length()==1){
            return s.length();
        }
        int max=0,curMax=0;
        int[] keyMap = new int[256];//测试的字符串都是英文字符，此处可以用一个256长度的数组来模拟hashmap
        for(int i=0;i<keyMap.length;i++){
            keyMap[i]=-1;
        }
        int m=0,n=0;
        while(n<s.length()){
            if(keyMap[s.charAt(n)]==-1){
                curMax=n-m+1;
            }else{
                if(keyMap[s.charAt(n)]>=m){//此处必须为>=,abba
                    m=keyMap[s.charAt(n)]+1;
                }
                curMax=n-m+1;
            }
            keyMap[s.charAt(n)]=n;
            n++;
            if(curMax>max){
                max=curMax;
            }
        }
        return max;
    }
}

Go代码

func lengthOfLongestSubstring(s string) int {
    if len(s)==0||len(s)==1{
        return len(s)
    }
    var i,j int
    var max int
    var keyMap map[byte]int=make(map[byte]int)
    var curMax int
    for j=0;j<len(s);{
        if _,ok:=keyMap[s[j]];!ok{
            curMax=j-i+1
            keyMap[s[j]]=j
            j++
            
        }else{
            if i<=keyMap[s[j]]{
                i=keyMap[s[j]]+1
            }
            
            keyMap[s[j]]=j
            curMax=j-i+1
            j++
        }
        if curMax>max{
            max=curMax
        }
    }
    return max
}