Word Abbreviation

摘要：链接注意第一个数字是的情况，这种也是不合法的。还有一个注意的就是要想和有相同的缩写，长度必须和它相同，所以只保留长度相同的。注意剪枝，当前长度已经超过就不需要继续了。二进制的做法是这样的，先对字典里面的单词进行处理。

链接：https://leetcode.com/problems...

注意第一个数字是0的情况，["a", "01"]这种也是不合法的。

public class Solution {
    public boolean validWordAbbreviation(String word, String abbr) {
        /* while loop, i for word, j for abbr
         * if it is number: count the number
         * i += number
         * else: compare word.charAt(i) == abbr.charAt(j)
         * end: i < len(word) && j < len(abbr)
         * return i == len(word) && j == len(abbr)
         */
         int i = 0, j = 0;
         while(i < word.length() && j < abbr.length()) {
             // character
             if(abbr.charAt(j) > "9" || abbr.charAt(j) <= "0") {
                 // characters not same
                 if(word.charAt(i) != abbr.charAt(j)) return false;
                 i++;  j++;
             }
             // count number
             else {
                 int count = 0;
                 while(j < abbr.length() && Character.isDigit(abbr.charAt(j))) {
                     count = 10 * count + (abbr.charAt(j) - "0");
                     j++;
                 }
                 i += count;
             }
         }
         
         return i == word.length() && j == abbr.length();
    }
}

        // if number: 
        // if character: check the same
        int i = 0;
        int m = word.length(), n = abbr.length();
        int count = 0;
        for(int j = 0; j < abbr.length(); j++) {
            char c = abbr.charAt(j);
            // number
            if(c >= "0" && c <= "9") {
                if(count == 0 && c == "0") return false;
                count = count * 10 + (c - "0");
            }
            // digit
            else {
                i += count;
                if(i >= word.length() || word.charAt(i) != c) return false;
                count = 0;
                i++;
            }
        }
        return i + count == m;

题目链接：https://leetcode.com/problems...

又是一道backtracking的题。看了这个博客的解法：
http://bookshadow.com/weblog/...

现在是穷举可能的结果，注意prune，然后check是否有和dict相同的。还有一个注意的就是要想和target有相同的缩写，长度必须和它相同，所以dict只保留长度相同的。注意剪枝，当前长度已经超过globalMin就不需要继续了。

public class Solution {
    public String minAbbreviation(String target, String[] dictionary) {
        // only keep the words has the same length
        int len = target.length();
        for(String s : dictionary) {
            if(s.length() == len) dict.add(s);
        }
        // no word has the same length as target
        if(dict.isEmpty()) return String.valueOf(target.length());
        globalMin = len;
        global = target;
        dfs(target, 0, 0, "");
        return global;
    }
    Set dict = new HashSet();
    int globalMin;
    String global;
    private void dfs(String target, int index, int len, String abbr) {
        // pruning
        if(len >= globalMin) return;
        // base case
        if(index == target.length()) {
            for(String word : dict) {
                if(validWordAbbreviation(word, abbr)) return;
            }
            globalMin = len;
            global = abbr;
            return;
        }
        // 2 subproblems:
        // 1. target[i] = char 
        // 2. target[i] = num
        dfs(target, index + 1, len + 1, abbr + target.charAt(index));
        int abbr_len = abbr.length();
        if(index == 0 || !Character.isDigit(abbr.charAt(abbr_len - 1))) {
            dfs(target, index + 1, len + 1, abbr + 1);
        }
        else {
            int num = 1 + (abbr.charAt(abbr_len - 1) - "0");
            dfs(target, index + 1, len, abbr.substring(0, abbr_len-1) + num);
        }
    }
    
    private boolean validWordAbbreviation(String word, String abbr) {
        // if number: 
        // if character: check the same
        int i = 0;
        int m = word.length(), n = abbr.length();
        int count = 0;
        for(int j = 0; j < abbr.length(); j++) {
            char c = abbr.charAt(j);
            // number
            if(c >= "0" && c <= "9") {
                if(count == 0 && c == "0") return false;
                count = count * 10 + (c - "0");
            }
            // digit
            else {
                i += count;
                if(i >= word.length() || word.charAt(i) != c) return false;
                count = 0;
                i++;
            }
        }
        return i + count == m;
    }
}

还有bit的方法，感觉好厉害！！完全没想出来。
二进制的做法是这样的，先对字典里面的单词进行处理。一个char一个char处理，如果该char和target对应位置上的一样，则记为1，否则记为0，这样处理完之后就知道哪些位置上的字母可以换成数字。对target进行缩写的时候，保留字母的记为1，换成数字的记为0，这样查target的abbr是否是word的缩写时，只需要把两者相与看是否和abbr相同即可。
我还是没搞懂这个到底是怎么想出来的，明天再看看。

public class Solution {
    public String minAbbreviation(String target, String[] dictionary) {
        len = target.length();
        globalMin = target.length()+1;
        global = 0;
        getBitDict(target, dictionary);
        // edge case: target in dict, no word with same len
        if(globalMin == 0) return target;
        if(dict.size() == 0) return String.valueOf(len);
        // backtracking
        dfs(0, 0, 0);
        return bitToString(target);
    }
    
    List dict = new ArrayList();
    int globalMin;
    int global;
    int len;
    private void dfs(int index, int curLen, int abbr) {
        // prune
        if(curLen >= globalMin) return;
        // base case
        if(index == len) {
            for(int word : dict) {
                if((word & abbr) == abbr) return;
            }
            globalMin = curLen;
            global = abbr;
            return;
        }
        // 1. character
        dfs(index + 1, curLen + 1, (abbr << 1) + 1);
        // 2. number
        if(index == 0 || (abbr%2) == 1) dfs(index + 1, curLen + 1, (abbr << 1));
        else dfs(index + 1, curLen, (abbr << 1));
    }

    
    private void getBitDict(String target, String[] dictionary) {
        // bit: 1. s[i] == target[i] => 1
        //      2. s[i] != target[i] => 0
        int len = target.length();
        for(String s : dictionary) {
            if(s.length() == len) {
                // edge case
                if(s.equals(target)) {
                    globalMin = 0;
                    return;
                }
                int bitString = 0;
                for(int i = 0; i < len; i++) {
                    bitString = bitString << 1;
                    if(target.charAt(i) == s.charAt(i)) bitString += 1;
                }
                dict.add(bitString);
            }
        }
    }
    
    private String bitToString(String target) {
        String result = "";
        int count = 0;
        for(int i = 0; i < len; i++) {
            if(((global >> len - i - 1) & 1) == 1) {
                if(count != 0) result += count;
                count = 0;
                result += target.charAt(i);
            }
            else count++;
        }
        if(count != 0) result += count;
        return result;
    }
}

练习白板第一天，板子买小了。思路写的也不整齐，擦了好几次，大概写了30分钟，还需要多多练习额。。dfs的题，下次写的时候，还是按照start -> arguments&return type -> base case -> subproblem的顺序来，pruning最后添上。明天研究性下怎么写时间复杂度的。