LeetCode 320. Generalized Abbreviation

原题链接在这里：https://leetcode.com/problems/generalized-abbreviation/

题目：

Write a function to generate the generalized abbreviations of a word.

Example:

Given word = "word", return the following list (order does not matter):

["word", "1ord", "w1rd", "wo1d", "wor1", "2rd", "w2d", "wo2", "1o1d", "1or1", "w1r1", "1o2", "2r1", "3d", "w3", "4"]

题解：

For DFS, the state needs word, current index, current string, res and count of abbreviated char.

对于当前char有两个选择: 第一缩写当前char, count+1, index+1. 第二不缩写当前char, 先append 非零的count再append当前char, count回零, index+1.

Backtracking时因为直接用string, 都是copy所以不担心.

Time Complexity: exponential. 在每一步递归时，有两个branches. 递归树全部展开会有2^n个叶子. 每一个叶子都相当于用了O(n)时长, 把StringBuilder变成String就用时O(n). n = word.length().

Space: O(n). n层stack, char array size也是n. regardless res.

AC Java:

class Solution {
    public List<String> generateAbbreviations(String word) {
        List<String> res = new ArrayList<String>();
        if(word == null || word.length() == 0){
            res.add("");
            return res;
        }
        
        dfs(word, 0, 0, "", res);
        return res;
    }
    
    private void dfs(String word, int i, int count, String cur, List<String> res){
        if(i == word.length()){
            if(count != 0){
                cur += count;
            }
            
            res.add(cur);
            return;
        }
        
        dfs(word, i+1, count+1, cur, res);
        
        if(count != 0){
            cur += count;
        }
        
        cur += word.charAt(i);
        dfs(word, i+1, 0, cur, res);
    }
}

AC Python:

class Solution:
    def generateAbbreviations(self, word: str) -> List[str]:
        res = []
        if len(word) == 0:
            res.append(word)
            return res
        
        self.dfs(word, 0, 0, "", res)
        return res
    
    def dfs(self, word: str, ind: int, count: int, item: str, res: List[str]) -> None:
        if ind == len(word):
            if count != 0:
                item += str(count)
            
            res.append(item)
            return
        
        self.dfs(word, ind + 1, count + 1, item, res)
        
        if count != 0:
            item += str(count)
        self.dfs(word, ind + 1, 0, item + word[ind], res)

Bit Manipulation 方法是用binary表示word的每一位. 0代表不缩写当前char, 1代表缩写当前char. 

word 若是用 0011表示就是不缩写wo, 缩写rd, 最后是wo2.

对于word的所有binary表示, 也就是0000到1111走一遍，每个binary变成string.

Time Complexity: O(n*2^n). n = word.length(), 一共有O(2^n)个binary表示. 每个binary变成string用时O(n).

Space: O(n), regardless res.

AC Java:

public class Solution {
    public List<String> generateAbbreviations(String word) {
        List<String> res = new ArrayList<String>();
        if(word == null){
            return res;
        }
        int n = 1<<word.length();
        char [] s = word.toCharArray();
        for(int i = 0; i<n; i++){
            res.add(abbr(i, s));
        }
        return res;
    }
    
    private String abbr(int num, char [] s){
        StringBuilder sb = new StringBuilder();
        int count = 0;
        for(int i = 0; i<s.length; i++, num>>=1){
            // 0 表示不缩写当前char, 1 表示缩写当前char
            if((num & 1) == 0){
                // 先加上之前的非零count, 再把count清零
                if(count != 0){
                    sb.append(count);
                    count = 0;
                }
                sb.append(s[i]);
            }else{
                count++;
            }
        }
        //最后的非零count不要忘记加进sb中
        if(count != 0){
            sb.append(count);
        }
        return sb.toString();
    }
}