[LintCode] Longest Substring Without Repeating Cha

Scliang 发布于2019-08-14 15:19 / 2436人阅读

摘要：哈希表法双指针法只有当也就是时上面的循环才会结束，，跳过这个之前的重复字符再将置为

Problem

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

Example

For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3.

For "bbbbb" the longest substring is "b", with the length of 1.

Note Solution

1. 哈希表法

HashMap Method I

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

HashMap Method II

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        if (s == null || s.length() == 0) return 0;
        HashMap map = new HashMap<>();
        int max = 1, start = 0;
        for (int i = 0; i < s.length(); i++) {
            Character cur = s.charAt(i);
            Integer rep = map.get(cur);
            if (rep != null && rep >= start) {
                max = Math.max(max, i-start);
                start = rep+1;
            }
            map.put(cur, i);
        }
        return Math.max(s.length()-start, max);
    }
}

2. 双指针法

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        int start = 0, end = 0, count = 0;
        boolean[] mark = new boolean[256];
        while (end < s.length()) {
            if (!mark[s.charAt(end)]) {
                mark[s.charAt(end)] = true;
                count = Math.max(count, end-start+1);
                end++;
            }
            else {
                while (mark[s.charAt(end)]) {
                    mark[s.charAt(start)] = false;
                    start++;
                }
                
                //只有当s.charAt(start) == s.charAt(end)
                //也就是mark[s.charAt(end)] == false时
                //上面的循环才会结束，start++，跳过这个之前的重复字符
                //再将mark[s.charAt(end)]置为true
                
                mark[s.charAt(end)] = true;
                end++;
            }
        }
        return count;
    }
}

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