LWC 66: 758. Bold Words in String

Problem:

Given a set of keywords words and a string S, make all appearances of all keywords in S bold. Any letters between and tags become bold. The returned string should use the least number of tags possible, and of course the tags should form a valid combination. For example, given that words = [“ab”, “bc”] and S = “aabcd”, we should return “aabcd”. Note that returning “aabcd” would use more tags, so it is incorrect.

Note:

words has length in range [0, 50].
words[i] has length in range [1, 10].
S has length in range [0, 500].
All characters in words[i] and S are lowercase letters.

思路： S的长度不长，所以可以用一个boolean数组记录每个位置是否需要加粗。起初，我for循环两次S，得到可能的subString判断是否在words中存在，结果超时了，解决超时的关键在于note，words的长度最多就10，所以干脆从可能的word中找是否有对应S的subString，这样即解决了超时。

Java版本如下：

    public String boldWords(String[] words, String S) {
        int n = S.length();
        boolean[] marked = new boolean[n];
        for (int i = 0; i < n; ++i) {
            for (String word : words) {
                if (i + word.length() <= n && word.equals(S.substring(i, i + word.length()))) {
                    for (int j = i; j < i + word.length(); ++j) {
                        marked[j] = true;
                    }
                }
            }
        }

        StringBuilder sb = new StringBuilder();
        int j = 0;
        while (j < n) {
            while (j < n && !marked[j]) {
                sb.append(S.charAt(j));
                j ++;
            }
            if (j >= n) break;
            StringBuilder tmp = new StringBuilder();
            while (j < n && marked[j]) {
                tmp.append(S.charAt(j));
                j ++;
            }
            if (tmp.length() != 0) sb.append("<b>" + tmp.toString() + "</b>");
        }
        return sb.toString();
    }

不过我们还可以采用IMOS累积法，这样可以省去一个for循环。

Java代码如下：

    public String boldWords(String[] words, String S) {
        int n = S.length();
        int[] hits = new int[n];
        for (int i = 0; i < n; ++i) {
            for (String word : words) {
                if (i + word.length() <= n && word.equals(S.substring(i, i + word.length()))) {
                    hits[i] ++;
                    hits[i + word.length()] --;
                }
            }
        }

        StringBuilder sb = new StringBuilder();
        for (int i = 1; i < n; ++i) hits[i] += hits[i - 1];
        for (int i = 0; i < n; ++i) {
            if (hits[i] > 0) {
                sb.append("<b>" + S.charAt(i) + "</b>");
            }
            else sb.append(S.charAt(i));
        }
        return sb.toString().replace("</b><b>", "");
    }

Python版本：

class Solution(object):
    def boldWords(self, words, S):
        """
        :type words: List[str]
        :type S: str
        :rtype: str
        """
        n = len(S)
        hits = [0] * (n + 1)
        for i in range(n):
            for word in words:
                if (i + len(word) <= n and word == S[i:i + len(word)]):
                    hits[i] += 1
                    hits[i + len(word)] -= 1
        for i in range(1, n + 1): hits[i] += hits[i - 1]
        ans = ""
        i = 0;
        while (i < n):
            while (i < n and hits[i] == 0):
                ans += S[i]
                i += 1
            if i >= n: break
            tmp = ""
            while (i < n and hits[i] != 0):
                tmp += S[i]
                i += 1
            if (len(tmp) != 0): ans += "<b>" + tmp + "</b>"
        return ans