【leedcode】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.

寻找最长的不重复串，略像滑动窗口。

char[] buff;
    //used
    int used ;
    public int lastIndexOf(char ch, int endIndex) {
            int i = used;
            for (; i >= endIndex; i--) {
                if (buff[i] == ch) {
                    return i;
                }
            }
            return -1;
    }
    public int lastIndexOf(char ch) {
        return lastIndexOf(ch, 0);
    }
    
     public int lengthOfLongestSubstring(String s) {
             if (s == null || s.isEmpty()) {
                 return 0;
             }
            int len = s.length();
            buff = new  char[len];
            buff[0] = s.charAt(0);
            used = 1;
            char t;
            int idx = -1;
            int top = 0;
            int first = 0;
            while(used<len){
                t = s.charAt(used);
                idx = lastIndexOf(t, first);
//                System.err.println("["+top+"]s=" + new String(buff) + "," + (idx>-1 ? ( "pos=" +  idx + " find " + t ) : "" ) + ",idx="  + used + ",first=" + first);
                
                if (idx > -1) {
                    top = Math.max(used - first, top);
                    first = idx+1;
                }
                
                buff[used] = t;
                used++;
            } //            System.err.println("["+top+"]s=" + new String(buff) + "," + (idx>-1 ? ( "pos=" +  idx + " find " + t ) : "" ) + ",idx="  + used + ",first=" + first);
            return Math.max(len - first, top);
        }

 上面这个粗鲁的代码，马马虎虎毕竟打败51 ~61%的code，lastIndexOf有优化的空间。

 不过答案真是美妙的活动窗口实现，赞！使用了字符作为坐标，索引作为值。

 public int lengthOfLongestSubstring(String s) {
        int n = s.length(), ans = 0;
        int[] index = new int[128]; // current index of character
        // try to extend the range [i, j]
        for (int j = 0, i = 0; j < n; j++) {
            i = Math.max(index[s.charAt(j)], i);
            ans = Math.max(ans, j - i + 1);
            index[s.charAt(j)] = j + 1;
        }
        return ans;
    }