320. Generalized Abbreviation
2017-01-24 15:01
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Write a function to generate the generalized abbreviations of a word.
Example:
Given word =
要枚举,要么递归要么迭代,递归算法思路:每个字母都只有两种情况,保留或者缩写隐藏。用一个数记录缩写的字母数,代码如下,虽然看着topsulotion写出来,仿佛已经理解,但还是感觉自己写写不出来:
public class Solution {
public List<String> generateAbbreviations(String word) {
List<String> result = new ArrayList<String>();
backtrack(result, new String(), word, 0, 0);
return result;
}
public void backtrack(List<String> list, String str, String word, int start, int count) {
if (start == word.length()) {
if (count > 0) {
str = str + count;
}
list.add(str);
} else {
backtrack(list, str, word, start + 1, count + 1);
backtrack(list, str + (count > 0? count: "") + word.charAt(start), word, start + 1, 0);
}
}
}
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"]
要枚举,要么递归要么迭代,递归算法思路:每个字母都只有两种情况,保留或者缩写隐藏。用一个数记录缩写的字母数,代码如下,虽然看着topsulotion写出来,仿佛已经理解,但还是感觉自己写写不出来:
public class Solution {
public List<String> generateAbbreviations(String word) {
List<String> result = new ArrayList<String>();
backtrack(result, new String(), word, 0, 0);
return result;
}
public void backtrack(List<String> list, String str, String word, int start, int count) {
if (start == word.length()) {
if (count > 0) {
str = str + count;
}
list.add(str);
} else {
backtrack(list, str, word, start + 1, count + 1);
backtrack(list, str + (count > 0? count: "") + word.charAt(start), word, start + 1, 0);
}
}
}
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