zoukankan      html  css  js  c++  java
  • LeetCode

    题目:

    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 substring is "b", with the length of 1.

    思路:

    1) 两个for循环,用map存访问过的字母,遇到相同的就停止,然后再从这个相同字母的下一个字母开始。超时了。

    package string;
    
    import java.util.HashMap;
    
    public class LongestSubstringWithoutRepeatingCharacters {
    
        public int lengthOfLongestSubstring(String s) {
            int len;
            if (s == null || (len = s.length()) == 0) return 0;
            HashMap<Character, Integer> map = new HashMap<Character, Integer>();
            int max = 0;
            for (int i = 0; i < len; ++i) {
                map.put(s.charAt(i), i);
                for (int j = i + 1; j < len; ++j) {
                    if (!map.containsKey(s.charAt(j))) {
                        map.put(s.charAt(j), j);
                    } else {
                        i = map.get(s.charAt(j));
                        break;
                    }
                }
                
                int count = map.size();
                if (count > max) {
                    max = count;
                }
                map.clear();
            }
            
            return max;
        }
        
        public static void main(String[] args) {
            // TODO Auto-generated method stub
            String s = "abcabcbb";
            LongestSubstringWithoutRepeatingCharacters l = new LongestSubstringWithoutRepeatingCharacters();
            System.out.println(l.lengthOfLongestSubstring(s));
        }
    
    }

    2)对方法一的简化,保持重复元素的上一个位置,往前挪动,不断计算

    package string;
    
    import java.util.HashMap;
    
    public class LongestSubstringWithoutRepeatingCharacters {
    
        public int lengthOfLongestSubstring(String s) {
            int len;
            if (s == null || (len = s.length()) == 0) return 0;
            HashMap<Character, Integer> map = new HashMap<Character, Integer>();
            int max = 0;
            int lastIndex = -1;
            for (int i = 0; i < len; ++i) {
                char c = s.charAt(i);
                if (map.containsKey(c) && lastIndex < map.get(c)) {
                    lastIndex = map.get(c);
                }
                
                if (i - lastIndex > max)
                    max = i - lastIndex;
                map.put(c, i);
            }
            
            return max;
        }
        
        public static void main(String[] args) {
            // TODO Auto-generated method stub
            String s = "abcabcbb";
            LongestSubstringWithoutRepeatingCharacters l = new LongestSubstringWithoutRepeatingCharacters();
            System.out.println(l.lengthOfLongestSubstring(s));
        }
    
    }
  • 相关阅读:
    hdu 2881 Jack's struggle(DP)
    [置顶] DataGridView控件---绑定数据方法
    Computational Geometry Template_Polygon
    Max retries exceeded with url
    Linux /proc 的意义
    [置顶] 生成树协议介绍
    StringBuffer与StringBuilder的异同
    linux 下Eclipse for C/C++的不常见设置
    [Elasticsearch] 分布式搜索
    oracle递归函数
  • 原文地址:https://www.cnblogs.com/null00/p/5029295.html
Copyright © 2011-2022 走看看