648. Replace Words 替换成为原来的单词

Posted 2020-12-31 immiao0319

［抄题］：

In English, we have a concept called root, which can be followed by some other words to form another longer word - let‘s call this word successor. For example, the root an, followed by other, which can form another word another.

Now, given a dictionary consisting of many roots and a sentence. You need to replace all the successor in the sentence with the root forming it. If a successor has many roots can form it, replace it with the root with the shortest length.

You need to output the sentence after the replacement.

Example 1:

Input: dict = ["cat", "bat", "rat"]
sentence = "the cattle was rattled by the battery"
Output: "the cat was rat by the bat"

 [暴力解法]：

时间分析：

空间分析：

 [优化后]：

时间分析：

空间分析：

［奇葩输出条件］：

［奇葩corner case］：

［思维问题］：

不知道怎么取单词前半截：.substring都忘了，实在是太弱了。

［英文数据结构或算法，为什么不用别的数据结构或算法］：

［一句话思路］：

［输入量］：空： 正常情况：特大：特小：程序里处理到的特殊情况：异常情况（不合法不合理的输入）：

［画图］：

［一刷］：

substring包左不包右，所以i需要 i <= word.length多加一位

［二刷］：

［三刷］：

［四刷］：

［五刷］：

  [五分钟肉眼debug的结果]：

［总结］：

取前半截用substring，而且右边界的i不包，要包就要加<=i的等号

［复杂度］：Time complexity: O(n) Space complexity: O(n)

［算法思想：迭代/递归/分治/贪心］：

［关键模板化代码］：

［其他解法］：

［Follow Up］：

［LC给出的题目变变变］：

 [代码风格] ：

 [是否头一次写此类driver funcion的代码] ：

 [潜台词] ：

class Solution {
    public String replaceWords(List<String> dict, String sentence) {
        //corner case
      if (dict == null || sentence.length() == 0) 
        return "";
      //initialization: set
      Set<String> set = new HashSet<String>();
      
      //put all the words into set
      for (int i = 0; i < dict.size(); i++) {
        set.add(dict.get(i));
      }
      
      //append the new result
      StringBuilder sb = new StringBuilder();
      String[] words = sentence.split("\s+");
      for (String word : words) {
        String prefix = "";
        for (int i = 1; i <= word.length(); i++) {
          prefix = word.substring(0, i);
          if (set.contains(prefix)) break;
        }
        sb.append(" " + prefix);
      }
      
      //return
      return sb.deleteCharAt(0).toString();
    }
}

View Code