leetcode Longest Substring Without Repeating Characters题解

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.

中文理解:给定一个字符串,求出字符串中不含重复字符的子序列的最大值。

解题思路:

解答一:

使用二重循环,外循环假设为初始的字符的下标,内循环从外循环开始,获取最大不重复子串的长度,若发现重复,跳出,与全局最大值比较,获得最大值。

代码(java):

class Solution {
    public int lengthOfLongestSubstring(String s) {
        if(s.length()<=1)return s.length();
        int max=-1;
        for(int i=0;i<s.length();i++){
            HashSet<Character> set=new HashSet<Character>();
            int unique=0;
            for(int j=i;j<s.length();j++){
                if(set.contains(s.charAt(j))){
                    break;
                }
                else{
                    unique++;
                    set.add(s.charAt(j));
                }
            }
            max=Math.max(max,unique);
        }
        return max;
    }
}

解法二:

使用两个指针来进行模拟,i,j两个指针,其中i指针在前,j指针在后,使用HashMap存储字符,下标,如果字符在子串中重复出现,则更新j值,同时每次修改max值来进行获取全局最大值。

代码(java):

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

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转载自blog.csdn.net/leo_weile/article/details/89474992