字符串中查找最大连续不重复字符串(Longest Substring Without Repeating Characters)__By leetCode

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

查找一个字符中最大连续不重复字符串

• One

public int lengthOfLongestSubstring(String s) {
        Set set = new HashSet<>();

        char[] charList = s.toCharArray();

        int length = 0;
        int sum =0;
        for (int i = 0,j=0; i < charList.length; i++) {
            if (set.contains(charList[i])){
                sum = set.size();
                set.clear();
                j++;
                i = j;
                set.add(charList[j]);
                length = Math.max(length , sum);
            }else {
                set.add(charList[i]);
                if (i == charList.length-1){
                    sum = set.size();
                    length = Math.max(length , sum);
                }
            }
        }

        return length;
    }

解析:
创建一个HashSet，存储对应字符，然后返回最大长度.

• Two

public int lengthOfLongest(String s){
        int n = s.length(), ans = 0;

        int[] index= new int[128];

        for (int current = 0,start=0; current < n; i++) {
            start = Math.max(index[s.charAt(current)],start);
            ans = Math.max(ans ,current-start+1);
            index[s.charAt(current)] = current+1;
        }
        return ans;
    }

解析:
用start标注重复字符的下一个起始位置，然后用current+1标注当前位置，current+1-start就是此次循环最大值。

• 备注
• int[26] for Letters ‘a’ - ‘z’ or ‘A’ - ‘Z’
• int[128] for ASCII
• int[256] for Extended ASCII