1. 题目
给定一个字符串和一个字符串字典,找到字典里面最长的字符串,该字符串可以通过删除给定字符串的某些字符来得到。
如果答案不止一个,返回长度最长且字典顺序最小的字符串。如果答案不存在,则返回空字符串。
示例 1:
输入:
s = "abpcplea", d = ["ale","apple","monkey","plea"]
输出:
"apple"示例 2:
输入:
s = "abpcplea", d = ["a","b","c"]
输出:
"a"说明:
所有输入的字符串只包含小写字母。
字典的大小不会超过 1000。
所有输入的字符串长度不会超过 1000。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/longest-word-in-dictionary-through-deleting
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
2. 解题
class Solution { //C++
public:string findLongestWord(string s, vector<string>& d) {string ans;int i, j, k;for(i = 0; i < d.size(); ++i){for(j = k = 0; j <s.size() && k<d[i].size(); ++j){if(s[j] == d[i][k])//匹配了,移动一位k++;}if(k == d[i].size())//都匹配过了{if(d[i].size() > ans.size())ans = d[i];else if(d[i].size() == ans.size() && d[i] < ans)ans = d[i];}}return ans;}
};
148 ms 14.8 MB
class Solution:# py3def findLongestWord(self, s: str, d: List[str]) -> str:ans = ""n = len(d)m = len(s)for i in range(n):j = k = 0while j < m and k < len(d[i]):if s[j]==d[i][k]:k += 1j += 1if k==len(d[i]):if (len(d[i]) > len(ans)) or (len(d[i])==len(ans) and d[i] < ans):ans = d[i]return ans
724 ms 16 MB