zoukankan      html  css  js  c++  java
  • 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.

    class Solution {
        public int lengthOfLongestSubstring(String s) {
            if(s == null){
                return 0;
            }
            char[] sc = s.toCharArray();
            if(sc.length > 1){
                return countLongestSubstring(sc);
            }
            return sc.length;
        }
        
        /**
         * counter longest substring 
         */
        private int countLongestSubstring(char[] sc){
            // current save substring list
            List<Character> currentSubstring = new ArrayList<>();
            // longest substring list
            List<Character> longestSubstring = new ArrayList<>();
            // flag index
            int flagIndex = 0;
            for (int i = 0;i < sc.length; i++) {
                char item = sc[i];
                if (!currentSubstring.contains(item)) {
                    // add to list
                    currentSubstring.add(item);
                } else {
                    if (longestSubstring.size() < currentSubstring.size()) {
                        // evalution value
                        longestSubstring = currentSubstring;
                    }
                    // judge index
                    if(flagIndex < sc.length -1){
                        i = flagIndex + 1;
                        flagIndex = i;
                        item = sc[i];
                    }
                    // init value
                    currentSubstring = new ArrayList<>();
                    currentSubstring.add(item);
                }
            }
            if (longestSubstring.size() < currentSubstring.size()) {
                // evalution value
                longestSubstring = currentSubstring;
            }
            // return result
            return longestSubstring.size();
        }
    }
    
    大道,在太极之上而不为高;在六极之下而不为深;先天地而不为久;长于上古而不为老
  • 相关阅读:
    Macaca拓展自己控件的方法
    pyparsing:定制自己的解析器
    利用 Traceview 精准定位启动时间测试的异常方法 (工具开源)
    Jenkins插件开发
    一个简陋的高并发请求脚本的演进过程
    python 自定义回调函数
    默认端口
    基本数据类型封装类比较
    grep命令和tail命令
    idea 拉取git新分支
  • 原文地址:https://www.cnblogs.com/GodBug/p/8495818.html
Copyright © 2011-2022 走看看