[Leetcode] 522. Longest Uncommon Subsequence II 解題報告

題目：

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].

思路：

思路還是沿着上一道題目的：我們首先對所有的字元串進行排序，不過讓長的字元串排在前面。然後依次掃描這些字元串，如果發現一個字元串隻出現了一次，并且不是它前面的所有字元串的子序列，那麼該字元串的長度就是所有字元串的最長非公共子序列了。

代碼：

class Solution {
public:
    int findLUSlength(vector<string>& strs) {
        unordered_map<string, int> hash;
        for(int i = 0; i < strs.size(); ++i) {
          ++hash[strs[i]];
        }
        vector<pair<string,int>> v;
        for(auto it = hash.begin(); it != hash.end(); ++it) {
            v.push_back(*it);
        }
        sort(v.begin(), v.end(), PairComp); // longer string comes first
        for(int i = 0; i < v.size(); ++i) {
           if(v[i].second == 1) {
               int j = 0;
               for(; j < i; ++j) {
                   if(isS1subsOfS2(v[i].first, v[j].first))
                     break;
               }
               if(j == i) {
                   return v[i].first.size();
               }
           }
        }
        return -1;
    }
private:
    struct PairCompare {
        bool operator() (pair<string,int> &a, pair<string,int> &b) {
            return a.first.size() > b.first.size();
        }
    } PairComp;
    bool isS1subsOfS2(string &s1, string &s2) {
        int j = 0, i = 0;
        for(; i < s1.size(); ++i) {
            while(j < s2.size() && s1[i] != s2[j]) {
                ++j;
            }
            if(j == s2.size()) {
               return false;
            }
            ++j;
        }
        return true;
    }
};