Given a string and a string dictionary, find the longest string in the dictionary that can be formed by deleting some characters of the given string. If there are more than one possible results, return the longest word with the smallest lexicographical order. If there is no possible result, return the empty string.
Example 1:
Input:
s = "abpcplea", d = ["ale","apple","monkey","plea"]
Output:
"apple"
Example 2:
Input:
s = "abpcplea", d = ["a","b","c"]
Output:
"a"
Note:
All the strings in the input will only contain lower-case letters.
The size of the dictionary won't exceed 1,000.
The length of all the strings in the input won't exceed 1,000.
1.思考
- 一开始就想到下面的方法,即一个个比较,最简单粗暴也最容易想到,最后的结果也反映了这个方法的不巧妙。
2.实现
- Runtime: 96ms( 41.61% )
- Memory: 29.6MB( 16.21% )
class Solution {
public:
string findLongestWord(string s, vector<string> d) {
string res;
for(auto dic:d){
if(cmp(s, dic)){//dic is in s
if(res.size()<dic.size() || //The length of dic is longer
(res.size()==dic.size() && res>dic)){
//smallest lexicographical order
res = dic;
}
}
}
return res;
}
bool cmp(string s, string d)
{
int ls = s.size();
int ld = d.size();
if(ls<ld)
return false;
int ps=0, count=0;
for(int i=0; i<ld; i++){
for(int j=ps; j<ls; j++){
if(d[i]==s[j]){
ps=j+1;//Mark s position
count++;//Calculate the number of same string
break;
}
if(j==ls-1)//If can't find the letter in s
return false;
}
}
if(count==ld)
return true;
else
return false;
}
};