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Copy pathAddBoldTagInString616.java
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Copy pathAddBoldTagInString616.java
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299 lines (272 loc) · 8.43 KB
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/**
* Given a string s and a list of strings dict, you need to add a closed pair
* of bold tag <b> and </b> to wrap the substrings in s that exist in dict. If
* two such substrings overlap, you need to wrap them together by only one pair
* of closed bold tag. Also, if two substrings wrapped by bold tags are
* consecutive, you need to combine them.
*
* Example 1:
*
* Input:
* s = "abcxyz123"
* dict = ["abc","123"]
* Output:
* "<b>abc</b>xyz<b>123</b>"
*
* Example 2:
* Input:
* s = "aaabbcc"
* dict = ["aaa","aab","bc"]
* Output:
* "<b>aaabbc</b>c"
*
* Note:
* The given dict won't contain duplicates, and its length won't exceed 100.
* All the strings in input have length in range [1, 1000].
*/
public class AddBoldTagInString616 {
public String addBoldTag(String s, String[] dict) {
Trie trie = constructTrie(dict);
StringBuilder sb = new StringBuilder();
int l = 0;
int r = 0;
char[] chars = s.toCharArray();
for (int i=0; i<s.length(); i++) {
int len = trie.search(chars, i);
if (len == 0) {
if (i >= r) {
if (r > l) {
sb.append("<b>");
sb.append(s.substring(l, r));
sb.append("</b>");
}
r = i + 1;
l = r;
sb.append(s.charAt(i));
}
} else {
r = Math.max(r, i+len);
}
}
if (r > l) {
sb.append("<b>");
sb.append(s.substring(l, r));
sb.append("</b>");
}
return sb.toString();
}
private Trie constructTrie(String[] dict) {
Trie res = new Trie();
for (String word: dict) {
res.addWord(word);
}
return res;
}
public String addBoldTag2(String s, String[] dict) {
StringBuilder sb = new StringBuilder();
int l = 0;
int r = 0;
for (int i=0; i<s.length(); i++) {
int len = 0;
for (String word: dict) {
if (s.startsWith(word, i) && word.length() > len) {
len = word.length();
}
}
if (len == 0) {
if (i >= r) {
if (r > l) {
sb.append("<b>");
sb.append(s.substring(l, r));
sb.append("</b>");
}
r = i + 1;
l = r;
sb.append(s.charAt(i));
}
} else {
r = Math.max(r, i+len);
}
}
if (r > l) {
sb.append("<b>");
sb.append(s.substring(l, r));
sb.append("</b>");
}
return sb.toString();
}
/**
* https://leetcode.com/problems/add-bold-tag-in-string/discuss/104262/short-java-solution
*/
public String addBoldTag3(String s, String[] dict) {
int n = s.length();
int[] mark = new int[n+1];
for(String d : dict) {
int i = -1;
while((i = s.indexOf(d, i+1)) >= 0) {
mark[i]++;
mark[i + d.length()]--;
}
}
StringBuilder sb = new StringBuilder();
int sum = 0;
for(int i = 0; i <= n; i++) {
int cur = sum + mark[i];
if (cur > 0 && sum == 0) sb.append("<b>");
if (cur == 0 && sum > 0) sb.append("</b>");
if (i == n) break;
sb.append(s.charAt(i));
sum = cur;
}
return sb.toString();
}
/**
* https://leetcode.com/problems/add-bold-tag-in-string/discuss/104263/Java-solution-Same-as-Merge-Interval.
*/
public String addBoldTag4(String s, String[] dict) {
List<Interval> intervals = new ArrayList<>();
for (String str : dict) {
int index = -1;
index = s.indexOf(str, index);
while (index != -1) {
intervals.add(new Interval(index, index + str.length()));
index +=1;
index = s.indexOf(str, index);
}
}
intervals = merge(intervals);
int prev = 0;
StringBuilder sb = new StringBuilder();
for (Interval interval : intervals) {
sb.append(s.substring(prev, interval.start));
sb.append("<b>");
sb.append(s.substring(interval.start, interval.end));
sb.append("</b>");
prev = interval.end;
}
if (prev < s.length()) {
sb.append(s.substring(prev));
}
return sb.toString();
}
class Interval {
int start, end;
public Interval(int s, int e) {
start = s;
end = e;
}
public String toString() {
return "[" + start + ", " + end + "]" ;
}
}
public List<Interval> merge(List<Interval> intervals) {
if (intervals == null || intervals.size() <= 1) {
return intervals;
}
Collections.sort(intervals, new Comparator<Interval>(){
public int compare(Interval a, Interval b) {
return a.start - b.start;
}
});
int start = intervals.get(0).start;
int end = intervals.get(0).end;
List<Interval> res = new ArrayList<>();
for (Interval i : intervals) {
if (i.start <= end) {
end = Math.max(end, i.end);
} else {
res.add(new Interval(start, end));
start = i.start;
end = i.end;
}
}
res.add(new Interval(start, end));
return res;
}
/**
* https://leetcode.com/problems/add-bold-tag-in-string/discuss/104262/short-java-solution
*/
public String addBoldTag5(String s, String[] dict) {
int n = s.length();
int[] mark = new int[n+1];
for(String d : dict) {
int i = -1;
while((i = s.indexOf(d, i+1)) >= 0) {
mark[i]++;
mark[i + d.length()]--;
}
}
StringBuilder sb = new StringBuilder();
int sum = 0;
for(int i = 0; i <= n; i++) {
int cur = sum + mark[i];
if (cur > 0 && sum == 0) sb.append("<b>");
if (cur == 0 && sum > 0) sb.append("</b>");
if (i == n) break;
sb.append(s.charAt(i));
sum = cur;
}
return sb.toString();
}
public String addBoldTag6(String s, String[] dict) {
if (s == null || s.length() == 0) return s;
StringBuilder sb = new StringBuilder();
int l = 0;
int r = 0;
int len = s.length();
for (int i=0; i<len; i++) {
for (String word: dict) {
if (s.startsWith(word, i)) {
r = Math.max(r, i+word.length());
}
}
if (i < r) continue;
if (l != r) {
sb.append("<b>").append(s.substring(l, r)).append("</b>");
}
sb.append(s.charAt(i));
l = i+1;
r = i+1;
}
if (l != r) {
sb.append("<b>").append(s.substring(l, r)).append("</b>");
}
return sb.toString();
}
}
class Trie {
Map<Character, Trie> children = new HashMap<>();
boolean end = false;
public void addWord(String word) {
addWord(word.toCharArray(), 0);
}
public void addWord(char[] chars, int i) {
if (i == chars.length) {
end = true;
return;
}
char c = chars[i];
if (children.containsKey(c)) {
children.get(c).addWord(chars, i+1);
} else {
Trie child = new Trie();
children.put(c, child);
child.addWord(chars, i+1);
}
}
public int search(char[] chars, int i) {
return search(chars, i, i, 0);
}
private int search(char[] chars, int i, int x, int len) {
if (x >= chars.length) return end ? x-i : len;
char c = chars[x];
if (!children.containsKey(c)) return end ? x-i : len;
return children.get(c).search(chars, i, x+1, len);
}
// public void print() {
// for (char c: children.keySet()) {
// System.out.println(c);
// children.get(c).print();
// }
// }
}
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