zoukankan      html  css  js  c++  java
  • LeetCode-3. Longest Substring Without Repeating Characters

    1.题目描述

    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.

    2.我的分析思路

    说实话,这题当时拿到没有什么思路,惭愧。。。

    3.其他的思路

    看了讨论之后,直接贴出代码吧,中间比较精妙。

    方法1:

    public class Solution {
        public int lengthOfLongestSubstring(String s) {
            int n = s.length();
            Set<Character> set = new HashSet<>();
            int ans = 0, i = 0, j = 0;
            while (i < n && j < n) {
                // try to extend the range [i, j]
                if (!set.contains(s.charAt(j))){
                    set.add(s.charAt(j++));
                    ans = Math.max(ans, j - i);
                } else {
                    set.remove(s.charAt(i++));
                }
            }
            return ans;
        }
    }
    

    方法2:

    public class Solution {
        public 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;
        }
    }
    
  • 相关阅读:
    C++PRIMER 阅读笔记 第三章
    一个for循环打印二维数组
    递归实现数组求和
    strlen 与 sizeof
    call,apply,bind,this
    js 原型继承
    vue 动画
    vuex学习心得
    vue+elementui dropdown 下拉菜单绑定方法
    vue 生命周期一点学习
  • 原文地址:https://www.cnblogs.com/f-zhao/p/6391757.html
Copyright © 2011-2022 走看看