383. Ransom Note
2017-05-21 11:15
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Given an arbitrary ransom note string and another string containing letters from all the magazines, write a function that will return true if the ransom note can be constructed from the magazines ; otherwise, it
will return false.
Each letter in the magazine string can only be used once in your ransom note.
Note:
You may assume that both strings contain only lowercase letters.
思路:
第一个想到的是将magazine的字母存入堆中,每次将note中的字母取出来找到magazine然后删除,找不到时返回false
但这样每次都会遍历一遍堆,所以采用牺牲空间,直接用map的方法,magazine映射到的单元加一,note的减一,这样时间效率变高
class Solution {
public:
vector<int> findAnagrams(string s, string p) {
vector<int> pv(256,0), sv(256,0), res;
if(s.size() < p.size())
return res;
for(int i = 0; i < p.size(); ++i)
{
++pv[p[i]];
++sv[s[i]];
}
if(pv == sv)
res.push_back(0);
for(int i = p.size(); i < s.size(); ++i)
{
++sv[s[i]];
--sv[s[i-p.size()]];
if(pv == sv)
res.push_back(i-p.size()+1);
}
return res;
}
};
will return false.
Each letter in the magazine string can only be used once in your ransom note.
Note:
You may assume that both strings contain only lowercase letters.
canConstruct("a", "b") -> false canConstruct("aa", "ab") -> false canConstruct("aa", "aab") -> true
思路:
第一个想到的是将magazine的字母存入堆中,每次将note中的字母取出来找到magazine然后删除,找不到时返回false
但这样每次都会遍历一遍堆,所以采用牺牲空间,直接用map的方法,magazine映射到的单元加一,note的减一,这样时间效率变高
class Solution {
public:
vector<int> findAnagrams(string s, string p) {
vector<int> pv(256,0), sv(256,0), res;
if(s.size() < p.size())
return res;
for(int i = 0; i < p.size(); ++i)
{
++pv[p[i]];
++sv[s[i]];
}
if(pv == sv)
res.push_back(0);
for(int i = p.size(); i < s.size(); ++i)
{
++sv[s[i]];
--sv[s[i-p.size()]];
if(pv == sv)
res.push_back(i-p.size()+1);
}
return res;
}
};
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