leetcode3. Longest Substring Without Repeating Characters

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本文介绍了两种寻找字符串中最长无重复字符子串的方法：暴力求解法与滑动窗口法，并提供了详细的Java代码实现。

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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.

解法一

暴力求法：两层循环，判断每次循环的最长长度，即为所求结果

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        if(s.length() == 0) {
            return 0;
        }
        int index = 0;
        int result = 0;
        HashMap<Character, Integer> map = new HashMap<Character, Integer>();
        for(int i = 0; i < s.length(); i++) {
            index = 0;
            for (int j = i; j < s.length(); j++) {
                if (!map.containsKey(s.charAt(j))) {
                    index++;
                } else {
                    map.clear();
                    break;
                }
                map.put(s.charAt(j), j);
                if (result < index) {
                    result = index;
                }
            }
        }
        return result;
    }
}

这里写图片描述

解法二

采用滑动窗口法，变量i依次遍历字符串，变量j记录当前循环中，最开始的字符的位置，如果当前字符在map中存在，则j取map中存在的字符的位置＋1和当前j的最大值，即Math.max(j, map.get(s.charAt(i)) + 1)，窗口的大小即为长度，最终取长度最大值。

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        if(s.length() == 0) {
            return 0;
        }
        int index = 0;
        int result = 0;
        HashMap<Character, Integer> map = new HashMap<Character, Integer>();
       for(int i = 0, j = 0; i < s.length(); i++) {
            if(map.containsKey(s.charAt(i))) {
                j = Math.max(j, map.get(s.charAt(i)) + 1);
                // j = map.get(s.charAt(i)) + 1; //abba
            }
            map.put(s.charAt(i), i);
            result = Math.max(result, i - j + 1);
        }
        return result;
    }
}