给你一个字符串 licensePlate
和一个字符串数组 words
,请你找出并返回 words
中的 最短补全词 。
补全词 是一个包含 licensePlate
中所有的字母的单词。在所有补全词中,最短的那个就是 最短补全词 。
在匹配 licensePlate
中的字母时:
- 忽略
licensePlate
中的 数字和空格 。 - 不区分大小写。
- 如果某个字母在
licensePlate
中出现不止一次,那么该字母在补全词中的出现次数应当一致或者更多。
例如:licensePlate
= "aBc 12c"
,那么它的补全词应当包含字母 'a'
、'b'
(忽略大写)和两个 'c'
。可能的 补全词 有 "abccdef"
、"caaacab"
以及 "cbca"
。
请你找出并返回 words
中的 最短补全词 。题目数据保证一定存在一个最短补全词。当有多个单词都符合最短补全词的匹配条件时取 words
中 最靠前的 那个。
示例 1:
输入:licensePlate = "1s3 PSt", words = ["step", "steps", "stripe", "stepple"] 输出:"steps" 解释:最短补全词应该包括 "s"、"p"、"s"(忽略大小写) 以及 "t"。 "step" 包含 "t"、"p",但只包含一个 "s",所以它不符合条件。 "steps" 包含 "t"、"p" 和两个 "s"。 "stripe" 缺一个 "s"。 "stepple" 缺一个 "s"。 因此,"steps" 是唯一一个包含所有字母的单词,也是本例的答案。
示例 2:
输入:licensePlate = "1s3 456", words = ["looks", "pest", "stew", "show"] 输出:"pest" 解释:licensePlate 只包含字母 "s" 。所有的单词都包含字母 "s" ,其中 "pest"、"stew"、和 "show" 三者最短。答案是 "pest" ,因为它是三个单词中在 words 里最靠前的那个。
示例 3:
输入:licensePlate = "Ah71752", words = ["suggest","letter","of","husband","easy","education","drug","prevent","writer","old"] 输出:"husband"
示例 4:
输入:licensePlate = "OgEu755", words = ["enough","these","play","wide","wonder","box","arrive","money","tax","thus"] 输出:"enough"
示例 5:
输入:licensePlate = "iMSlpe4", words = ["claim","consumer","student","camera","public","never","wonder","simple","thought","use"] 输出:"simple"
提示:
1 <= licensePlate.length <= 7
licensePlate
由数字、大小写字母或空格' '
组成1 <= words.length <= 1000
1 <= words[i].length <= 15
words[i]
由小写英文字母组成
Related Topics
哈希
/**
* 哈希
* @param licensePlate
* @param words
* @return
*/
public static String shortestCompletingWord(String licensePlate, String[] words) {
if (licensePlate == null || words == null || words.length == 0) return "";
int[] chNums = new int[26];
for (char c : licensePlate.toCharArray()) {
if (c >= 'A' && c <= 'Z') c += ('a' - 'A');
if (c >= 'a' && c <= 'z') {
chNums[c - 'a']++;
}
}
int idx = -1, len = Integer.MAX_VALUE;
for (int i = 0; i < words.length; i++) {
boolean flag = true;
int[] chTmpNums = new int[26];
for (char c : words[i].toCharArray()) {
if (c >= 'A' && c <= 'Z') c += ('a' - 'A');
if ((c >= 'a' && c <= 'z')) {
chTmpNums[c - 'a']++;
}
}
for (int j = 0; j < 26; j++) {
if (chTmpNums[j] < chNums[j]) {
flag = false;
break;
}
}
if (flag && len > words[i].length()) {
idx = i;
len = words[i].length();
}
}
return idx == -1 ? "" : words[idx];
}
测试用例
public static void main(String[] args) {
String licensePlate = "1s3 PSt";
String[] words = new String[]{"step", "steps", "stripe", "stepple"};
String ansStr = ShortestCompletingWord.shortestCompletingWord(licensePlate, words);
System.out.println("ShortestCompletingWord demo01 result : " + ansStr);
licensePlate = "1s3 456";
words = new String[]{"looks", "pest", "stew", "show"};
ansStr = ShortestCompletingWord.shortestCompletingWord(licensePlate, words);
System.out.println("ShortestCompletingWord demo02 result : " + ansStr);
licensePlate = "Ah71752";
words = new String[]{"suggest", "letter", "of", "husband", "easy", "education", "drug", "prevent", "writer", "old"};
ansStr = ShortestCompletingWord.shortestCompletingWord(licensePlate, words);
System.out.println("ShortestCompletingWord demo03 result : " + ansStr);
licensePlate = "OgEu755";
words = new String[]{"enough", "these", "play", "wide", "wonder", "box", "arrive", "money", "tax", "thus"};
ansStr = ShortestCompletingWord.shortestCompletingWord(licensePlate, words);
System.out.println("ShortestCompletingWord demo04 result : " + ansStr);
licensePlate = "iMSlpe4";
words = new String[]{"claim", "consumer", "student", "camera", "public", "never", "wonder", "simple", "thought", "use"};
ansStr = ShortestCompletingWord.shortestCompletingWord(licensePlate, words);
System.out.println("ShortestCompletingWord demo05 result : " + ansStr);
}
测试结果
ShortestCompletingWord demo01 result : steps
ShortestCompletingWord demo02 result : pest
ShortestCompletingWord demo03 result : husband
ShortestCompletingWord demo04 result : enough
ShortestCompletingWord demo05 result : simple