Substring with Concatenation of All Words
Feb 24 '12
5071 / 18608
You are given a string, S, and a list of words, L, that are all of the same length. Find all starting indices of substring(s) in S that is a concatenation of each word in L exactly once and without any intervening characters.
For example, given:
S: "barfoothefoobarman"
L: ["foo", "bar"]
You should return the indices: [0,9]
.
(order does not matter).
注:未通过所有case
“aaa”, [a, b],这个case我自己返回是空,但是在leetcode上返回(0,1)。让我很郁闷。
假如unordered_map可以算作是O(1)的话,则这个算法应该是O(n),n = S.size();
class Solution {
public:
unordered_map<string, int> dic;
vector<int> t;
void load(const string& s, const vector<string> &l) {
int n = l.size();
int cnt = 0;
for (int i = 0; i < n; ++i)
{
auto it = dic.find(l[i]);
if (it == dic.end()) {
dic[l[i]] = cnt++;
t.push_back(1);
}
else {
t[it->second]++;
}
}
}
vector<int> findSubstring(string S, vector<string> &L) {
vector<int> res;
int n = L.size();
int len = S.size();
int K = L[0].size();
load(S, L);
vector<int> v(len);
for (int i = 0; i < len - K; ++i)
{
auto it = dic.find(S.substr(i, K));
if (it == dic.end()) v[i] = -1;
else v[i] = it->second;
}
int total = 0, last = len - K * n;
vector<bool> mk(len, false);
res.clear();
for (int i = 0; i <= last; ++i)
{
if (mk[i]) continue;
int start = i, end, cur;
while (start <= last && dic.find(S.substr(start, K)) == dic.end())
mk[start] = true,start += K;
if (start > last) continue;
if (n == 1) {
res.push_back(start);
mk[start] = true;
continue;
}
end = start + K;
vector<int> cnt(t);
total = 1;
auto it = dic.find(S.substr(start, K));
cnt[it->second]--;
while (end < len) {
auto ie = dic.find(S.substr(end, K));
if (ie == dic.end()) {
while (start <= end) mk[start] = true, start += K;
break;
}
else {
cur = ie->second;
if (cnt[cur] == 0) {
while (start <= end) {
mk[start] = true;
cnt[v[start]]++;
total--;
if (cur == v[start]) break;
start += K;
}
}
cnt[cur]--;
total ++;
if (total == n) {
res.push_back(start);
cnt[v[start]]++;
total--;
mk[start]=true;
start += K;
}
end += K;
}
}
}
return res;
}
};