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Leetcode 30. 串联所有单词的子串
2022-07-06 11:34:00 【Java全栈研发大联盟】
代码如下:
public static List<Integer> findSubstring(String s, String[] words) {
List<Integer> result = new ArrayList<>();
int length = 0;
for (String word : words) {
length += word.length();
}
for (int i = 0; i <= s.length() - length; i++) {
String substring = s.substring(i, i + length);
boolean tran = tran(substring, words);
if (tran) {
result.add(i);
}
}
return result;
}
public static boolean tran(String str, String[] words) {
//处理边界情况
if (words.length == 1) {
return str.equals(words[0]);
}
boolean result = false;
for (int i = 0; i < words.length; i++) {
if (str.startsWith(words[i])) {
StringBuilder stringBuilder = new StringBuilder(str);
String substring = stringBuilder.delete(0, words[i].length()).toString();
String[] restWords = subWords(words, i);
if (tran(substring, restWords)) {
result = true;
}
}
}
return result;
}
public static String[] subWords(String[] words, int i) {
String[] strings = Arrays.copyOf(words, words.length);
for (int k = i; k < strings.length - 1; k++) {
strings[k] = strings[k + 1];
}
strings = Arrays.copyOfRange(strings, 0, strings.length - 1);
return strings;
}
但是还是有一部分变态的输入情况,导致代码超时。 太无耻了这种输入情况
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