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。
Valid Word Abbreviation
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。
</>复制代码
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();
}
}
</>复制代码
// 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;
Minimum Unique Word Abbreviation
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。
</>复制代码
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。
</>复制代码
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。
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