Leetcode 3 无重复字符的最长子串(中等）（滑动窗口）

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分类专栏：笔记文章标签： leetcode java

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该博客介绍了如何使用滑动窗口算法来寻找字符串中不包含重复字符的最长子串长度。在Java实现中，通过HashMap存储字符及其索引，不断更新最大子串长度。当遇到重复字符时，调整窗口左边界。博客深入解析了HashMap在算法中的应用和滑动窗口的原理。

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给定一个字符串 s ，请你找出其中不含有重复字符的最长子串的长度。

示例 1:

输入: s = “abcabcbb”
输出: 3
解释: 因为无重复字符的最长子串是 “abc”，所以其长度为 3。

package com.xuyinda;

    import java.util.HashMap;
    import java.util.List;

    public class Solution {

            public static int lengthOfLongestSubstring(String s) {
                if(s==null){
                    return 0;
                }
                HashMap<Character, Integer> map = new HashMap<Character, Integer>();
                int left = 0;
                int max = 0;
                for(int i = 0; i <s.length();i++){
                    if(map.containsKey(s.charAt(i))) {
                    left = Math.max(left,map.get(s.charAt(i))+1);
                    }
                    //解释点一
                    map.put(s.charAt(i),i);
                    max = Math.max(max,i - left + 1);
                    //解释点二
                }
                return max;


            }

    }

Explanation 2 ：It is easy to understand this part. (i-left+1) means the current window’s length. And the variable max represents the longest window’s length.
If the currrent window’s length is bigger than the last max, the value of max will be replaced by the bigger one.

Explanation 1 ：If the new character repeats with the character in the map, the window’s left border will change. At first, I thought left = Math.max(map.get(s.charAt(i))+1) was enough. However, I missed one situation that the biggest index number of the character in the map might be smaller than the index number of current left border.

Summary：By resolving this question, I get a deep understanding of Hash Map and the algorithm of sliding window.