LeetCode Top 100 Liked Questions 3. Longest Substring Without Repeating Characters(Java版； Hard)

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LeetCode Top 100 Liked Questions 3. Longest Substring Without Repeating Characters(Java版; Medium)

题目描述

Given a string, find the length of the longest substring without repeating characters.

Example 1:

Input: "abcabcbb"
Output: 3 
Explanation: The answer is "abc", with the length of 3. 
Example 2:

Input: "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.
Example 3:

Input: "pwwkew"
Output: 3
Explanation: 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) {
        int n = s.length();
        if(n<=1){
            return n;
        }
        int left=0, right=0;
        int max = 1;
        Map<Character,Integer> map = new HashMap<>();
        boolean flag = false;
        while(right<n){
            char ch = s.charAt(right);
            if(map.containsKey(ch)){
                // 注意:map.get(ch)+1可能比当前的left小! 
                left = Math.max(left, map.get(ch)+1);
            }
            map.put(ch, right);
            
            max = Math.max(max, right - left + 1);
            right++;
        }
        return max;
    }
}

class Solution {
    public int lengthOfLongestSubstring(String s) {
        int n = s.length();
        if(n<=1){
            return n;
        }
        int left=0, right=0;
        int max = 1;
        Map<Character,Integer> map = new HashMap<>();
        boolean flag = false;
        while(right<n){
            char ch = s.charAt(right);
            map.put(ch, map.getOrDefault(ch, 0)+1);
            if(map.get(ch)==2){
                while(map.get(ch)==2){
                    char ch2 = s.charAt(left);
                    map.put(ch2, map.get(ch2)-1);
                    left++;
                }
            }
            max = Math.max(max, right - left + 1);
            right++;
        }
        return max;
    }
}

第一次做, 更新左边界的while循环最重要, 在寻找和右边界元素相同的元素时, 需要把沿途碰到的出现过的元素的出现次数重新赋值为0; 这题有点类似剑指offer面试题42连续子数组的最大和, 不过这两道题更新左边界的思想是不同的

class Solution {
    public int lengthOfLongestSubstring(String s) {
        if(s==null || s.length()==0)
            return 0;
        int left=0;
        int[] arr= new int[128];
        int max=0;
        for(int right=0; right<s.length(); right++){
            if(arr[s.charAt(right)]==0){
                arr[s.charAt(right)]++;
                max = Math.max(max, right - left + 1);
            }
            else{ //arr[s.charAt(right)]==1
                //找到重复的元素
                while(s.charAt(left)!=s.charAt(right)){
                    //寻找和右边界元素相同的元素时, 沿途碰到的其他出现过的元素需要把出现次数减为0
                    if(arr[s.charAt(left)]==1)
                        arr[s.charAt(left)]--;
                    left++;
                }
                //更新左边界
                left++;
            }
        }
        return max;
    }
}

题解中的答案, 和我的想法一致, 但是做法简介; 使用哈希表记录出现过的元素及对应的索引, 之后之后再遍历到该元素, 需要更新左边界left吗? 不一定, 如果哈希表中该元素对应的索引小于left就不用更新left, 如果哈希表中该元素对应的索引大于等于left, 就得更新left, 代码中用left=Math.max()实现这个想法, 比我的while循环快

• “abctbefa” 第二次碰到a的时候, 由于此时左边界left指向c, 哈希表中a的索引小于left, 所以不用更新left. 注意这之后得接着更新a在哈希表中的索引, 体现了execute → update 处理流程

class Solution {
    public int lengthOfLongestSubstring(String s) {
        if (s.length()==0) return 0;
        HashMap<Character, Integer> map = new HashMap<Character, Integer>();
        int max = 0;
        int left = 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);
            }
            map.put(s.charAt(i),i);
            max = Math.max(max,i-left+1);
        }
        return max;
        
    }
}