Monday, August 8, 2016

Google – Compress String II


Google – Compress String II
假设一个文件压缩后的表示是:
#3, #5, #6, 2 5, #8
#d表示单个数字d
两个数字"i j"表示取前i个数字,组长度为j的字符串。
比如 "#3" 就表示3
"2 5"表示前面两个数字,就是56,组成长度为5的字符串,就是56565
所以上面的例子,解压缩之后应该是: 3 5 6 5 6 5 6 5 8
要求是写一个压缩算法,把string按要求压缩。
比如 2 3 4 5 2 3 4 5 1, 返回#2, #3, #4, #5, 4 4, #1
[Solution]
一边扫描一边用hashtable 记录window size为2的pattern, map到这个Pattern所有出现过的位置,如果发现重复的,就开始往后match。
面经上写着window size为3,并且map到最后出现的位置,应该是不对。首先window size为3就不知道是为什么。其次map到最后出现的位置也有问题,比如
3 5 6 1 4 5 6 1 5 5 6 1 4 5 6 1 5
第二个5 6 1后面是5,并不能match第一个5 6 1后面的4。要直到第三个5 6 1才能match前面8个字符。但是如果5 6 1map到最后出现的位置的话,是无法继续match下去的。
  public String compress(String s) {
    if (s == null || s.isEmpty()) {
      return s;
    }
 
    Map<String, List<Integer>> map = new HashMap<>();
    StringBuilder sb = new StringBuilder();
    sb.append("#" + s.charAt(0));
    for (int i = 1; i < s.length(); i++) {
      String pattern = String.format("%c%c", s.charAt(i - 1), s.charAt(i));
      if (!map.containsKey(pattern)) {
        map.put(pattern, new ArrayList<>());
        sb.append(" ").append("#" + s.charAt(i));
      }
      else {
        boolean found = false;
        for (int j : map.get(pattern)) {
          int l = j + 1;
          int r = i + 1;
          while (r < s.length() && s.charAt(l) == s.charAt(r)) {
            l++;
            r++;
          }
          if (l < i - 1) {
            continue;
          }
          else {
            int prevNum = i - j;
            int len = r - i + 1;
            sb.delete(sb.length() - 3, sb.length());
            sb.append(" ").append(prevNum + " " + len);
            found = true;
            i = r - 1;
            break;
          }
        }
        if (!found) {
          sb.append(" ").append("#" + s.charAt(i));
        }
      }
 
      map.get(pattern).add(i);
    }
 
    return sb.toString();
  }
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