【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 .

Given "bbbbb", the answer is "b", with the length of .

Given "pwwkew", the answer is "wke", with the length of . 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;
    }