LeetCode 616. Add Bold Tag in String

原题连接在这里:https://leetcode.com/problems/add-bold-tag-in-string/description/html

题目:app

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.post

Example 1:ui

Input: 
s = "abcxyz123"
dict = ["abc","123"]
Output:
"<b>abc</b>xyz<b>123</b>"

Example 2:url

Input: 
s = "aaabbcc"
dict = ["aaa","aab","bc"]
Output:
"<b>aaabbc</b>c"

Note:spa

  1. The given dict won't contain duplicates, and its length won't exceed 100.
  2. All the strings in input have length in range [1, 1000].

题解:code

相似Merge Intervals. 标记出dict中每一个word所在s的起始结束位置. sort后merge.htm

或者直接用boolean array来标记s的当前char是否出如今dict中word所在s的substring内.blog

Time Complexity: O(dict.length*s.length()*x). x是dict中word的平均长度.ip

Space: O(s.length()).

AC Java:

 1 class Solution {
 2     public String addBoldTag(String s, String[] dict) {
 3         if(s == null || s.length() == 0 || dict == null || dict.length == 0){
 4             return s;
 5         }
 6         
 7         boolean [] mark = new boolean[s.length()];
 8         for(String word: dict){
 9             for(int i = 0; i<=s.length()-word.length(); i++){
10                 if(s.startsWith(word, i)){
11                     Arrays.fill(mark, i, i+word.length(), true);
12                 }
13             }
14         }
15         
16         int i = 0;
17         StringBuilder sb = new StringBuilder();
18         while(i<mark.length){
19             if(mark[i]){
20                 sb.append("<b>");
21                 while(i<mark.length && mark[i]){
22                     sb.append(s.charAt(i++));
23                 }
24                 
25                 sb.append("</b>");
26             }else{
27                 sb.append(s.charAt(i++));
28             }
29         }
30         
31         return sb.toString();
32     }
33 }
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