剑指offer_最长不含重复字符的子字符串 

 1 import java.util.Arrays;
 2 
 3 public class Solution {
 4 
 5     public int longestSubStringWithoutDuplication(String str) {
 6         int curLen=0;
 7         int maxLen=0;
 8         int [] index =new int[26];
 9         Arrays.fill(index, -1);
10         for (int i = 0; i < str.length(); i++) {
11             int preindex = index[str.charAt(i)-'a'];
12             if(preindex==-1||i-preindex>curLen){
13                 curLen++;
14             }
15             else{
16                 maxLen=Math.max(curLen, maxLen);
17                 curLen=i-preindex;
18             }
19             index[str.charAt(i)-'a']=i;
20         }
21         maxLen=Math.max(curLen, maxLen);
22         return maxLen;
23         
24     }
25     
26     public static void main(String[] args) {
27         Solution s = new Solution();
28         int longestSubStringWithoutDuplication = s.longestSubStringWithoutDuplication("arabcacf");
29         System.out.println(longestSubStringWithoutDuplication);
30     }
31 }

1、若第i个字符在之前没出现过，则 f(i) = f(i-1) + 1;

2、若第i个字符在之前出现过，

计算第i个字符距离上次出现之间的距离为d

(a)若d <= f(i-1)，则说明第i个字符上次出现在f(i-1)对应的不重复字符串之内，那么这时候更新 f(i) = d

(b)若d > f(i-1)，则无影响,f(i) = f(i-1) + 1

posted @ 2019-09-08 11:11  chyblogs  阅读(100)  评论(0)    收藏  举报