[LeetCode] 522. Longest Uncommon Subsequence II

轉自LeetCode

Given a list of strings, you need to find the longest uncommon subsequence among them. The longest uncommon subsequence is defined as the longest subsequence of one of these strings and this subsequence should not be any subsequence of the other strings.

A subsequence is a sequence that can be derived from one sequence by deleting some characters without changing the order of the remaining elements. Trivially, any string is a subsequence of itself and an empty string is a subsequence of any string.

The input will be a list of strings, and the output needs to be the length of the longest uncommon subsequence. If the longest uncommon subsequence doesn't exist, return -1.

Example 1:

Input: "aba", "cdc", "eae"
Output: 3

Note:

1. All the given strings' lengths will not exceed 10.
2. The length of the given list will be in the range of [2, 50].

<Solution>

521. Longest Uncommon Subsequence I 的衍生題

想法如下

• 由 521 的經驗，只要長度最長，且沒有其他字串一樣長，那就會是答案。因此，先將字串根據長度，做降冪排列

• 從最長的字串開始檢查，只要當前這個字串，不是其餘字串的子字串，那就是答案了。這邊要特別注意的是子字串的定義，根據題目 aa 是 aba 的子字串，因為只要將 aba 裡面的 b 給刪除(兩個 a 的順序並沒有變)，就可以組出 aa，所以檢查方式要特別寫

code 如下

C++

	class Solution {
	public:
	int findLUSlength(vector<string>& strs) {
	//>> sort strings in descending order
	sort(strs.begin(), strs.end(), [](string a, string b) {
	return a.length() == b.length() ? a > b : a.length() > b.length();
	});
	const int LEN = strs.size();
	const int UPPER_BOUND = LEN - 1;
	bool found;
	for(int i = 0; i < LEN; i++) {
	if(i == UPPER_BOUND || strs[i].compare(strs[i+1]) != 0) {
	found = true;
	for(int j = 0; j < LEN; j++) {
	if(j != i && isSubStr(strs[i], strs[j])) {
	found = false;
	break;
	}
	}
	if(found) {
	return strs[i].length();
	}
	}
	}
	return -1;
	}
	//>> check string a is the substr of string b or not
	bool isSubStr(const string &a, const string &b) {
	if(a.compare(b) == 0) {
	return true;
	}
	const int LEN = a.length();
	int i = 0;
	for(const auto &c : b) {
	if(i < LEN && a[i] == c) {
	++i;
	}
	}
	return i == LEN;
	}
	};

view raw longestUncommonSubStr2.cpp hosted with ❤ by GitHub

Java

	class Solution {
	public int findLUSlength(String[] strs) {
	//>> decending order
	Arrays.sort(strs, new Comparator<String>() {
	public int compare(String a, String b) {
	return b.length() - a.length();
	}
	});
	final int upper_bound = strs.length - 1;
	boolean found;
	for(int i = 0; i < strs.length; i++) {
	if(i == upper_bound || !strs[i].equals(strs[i+1])) {
	found = true;
	for(int j = 0; j < strs.length; j++) {
	if(j != i && isSubStr(strs[i], strs[j])) {
	found = false;
	break;
	}
	}
	if(found == true) {
	return strs[i].length();
	}
	}
	}
	return -1;
	}
	//>> chech a is substr of b or not
	private boolean isSubStr(final String a, final String b) {
	if(a.equals(b)) {
	return true;
	}
	int i = 0;
	final int len = a.length();
	for(char c : b.toCharArray()) {
	if(i < len && a.charAt(i) == c) {
	++i;
	}
	}
	return i == a.length();
	}
	}

view raw longestUncommonSubStr2.java hosted with ❤ by GitHub