LeetCode:3.Longest Substring Without Repeating Characters

思路:看到题目首先想到最大字符串匹配KMP算法

 1 public static int lengthOfLongestSubstring(String s) {
 2         int maxLength = 0;
 3         StringBuilder sb = new StringBuilder(s);
 4         a:for(int i = 0;i<sb.length();i++){
 5             StringBuilder sb2 = new StringBuilder("");
 6             sb2.append(sb.substring(i,i+1));
 7             b:for(int j=i+1;j<sb.length();j++){
 8                 c:for(int k =0;k<sb2.length();k++){
 9                     if(!sb.substring(j,j+1).equals(sb2.substring(k,k+1))){
10
11                     }else{
12                         if(maxLength<j-i)
13                             maxLength = j-i;
14                         break b;
15                     }
16                     if(k == sb2.length())
17                         sb2.append(sb.substring(j,j+1));
18                 }
19             }
20         }
21         return maxLength;
22 }

参考后代码

public static int lengthOfLongestSubstring(String s) {
        int n = s.length(), ans = 0;
        Map<Character, Integer> map = new HashMap<>(); // current index of character
        // try to extend the range [i, j]
        for (int j = 0, i = 0; j < n; j++) {
            if (map.containsKey(s.charAt(j))){
                i = Math.max(map.get(s.charAt(j)), i);
            }
            ans = Math.max(ans, j - i + 1);
            map.put(s.charAt(j), j + 1);
        }
        return ans;
    }
时间: 2024-08-29 15:57:56

LeetCode:3.Longest Substring Without Repeating Characters的相关文章

Python 解leetcode:3. Longest Substring Without Repeating Characters

题目描述:求一个字符串的不含重复字符的最长连续子串的长度: 思路: 使用一个哈希表保存字符出现的位置: 使用left和right分别表示子串的最左和最右字符的下标: 遍历字符串,如果当前字符在哈希表中并且当前字符在哈希中的位置大于left,表明left和right之间存在和right索引上相同的字符,此时把left置为当前值在哈希表中的值: 把当前索引值+1,表示是第几个字符,求出当前最大长度: 3-4步重复,直到获得遍历完成: 感觉这是个动态规划题,暂时没用动态规划分析,后续再说. class

LeetCode解题思路: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.

【LeetCode】003 Longest Substring Without Repeating Characters

题目:LeetCode 003 Longest Substring Without Repeating Characters 题意:给一个字符串,找到一个没有重复字符的最长子串. 样例:”bbbbb” 返回”b”的长度1:”abcabcbb”返回”abc”的长度3. 思路: 动态规划.dp[i]表示以第i个字符为结尾的无重复字符的子串的最长的长度,需要一个辅助数组idx[s[i]]记录字符当前字符s[i]上一次出现的位置,如果未出现过则为-1.所以得到如下的递推公式: 另外由于不确定字符串长度的

LeetCode Problem 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.

LeetCode No.3 Longest Substring Without Repeating Characters

Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3. For "bbbbb" the longest subst

【LeetCode】3. Longest Substring Without Repeating Characters

题目: 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

[Leetcode 3, Medium] Longest Substring Without Repeating Characters

Problem: Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3. For "bbbbb" the long

Leetcode OJ #3 Longest Substring Without Repeating Characters

https://leetcode.com/problems/longest-substring-without-repeating-characters/ 题目: 找一个字符串的连续子串,使得该子串里不含相同字母,求符合条件的最长的子串的长度. 算法: DP----后缀思想 最初考虑过用二分答案,发现复杂度应该是O(n*n*log(n)),其中n*n验证一个答案对不对,log(n)是枚举可能答案的长度.然后就放弃,考虑DP dp[i]:以str[i]结尾的符合条件的子串的最大长度. 那么dp[i

LeetCode题解 #3 Longest Substring Without Repeating Characters

找出字符串中没有相同字符的的最长串 注意这里的 Characters指的是字符,不是字母,就是说|/?~这样的字符都会出现,所以要用到ASCII码 最简单的方法是,从第一个字符开始,往后一个个判断,里面有没有重复的字符,如果重复了则记录下长度. 例如:abcabcbb 第一次:abc 重复于a  长度3 第二次:bca 重复与b  长度3 第三次:cab 重复与c  长度3 ...... 但这种方法很耗时 如果是 abcdefghijk这种 第一次就找到了的 abcdefghijk 但还要第二次