1. 题目
You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.
For example, given:
s: "barfoothefoobarman"
words: ["foo", "bar"]
You should return the indices: [0,9].
(order does not matter).
2. 思路
没有特别的好方法,相当于死扛计算的。
注意:words里可能有重复。
注意:“ab”,“ba”, "aa", "bb"这种,彼此交叉的。
遍历起点,每次找到一个匹配点,然后验证是否符合需求。然后+1从下一个点开始找下一个起点。
起点当前的点的第一个串是其中一个。
验证方法是,看是否能匹配到words.size个。特殊处理words里重复的情况,通过计数处理。
为每一个起点开始的匹配,为word命名一个version,记录下version。重复出现version时进行count比较。count超了就失败。
3. 代码
耗时:849ms
class Solution {
public:
vector<int> findSubstring(string& s, vector<string>& words) {
//cout << "s=" << s << endl;
vector<int> ret;
if (words.size() == 0) {
return ret;
}
unordered_map<string, int> map;
unordered_map<string, int> freq;
for (int i = 0; i < words.size(); i++) {
map[words[i]] = 0;
if (freq.find(words[i]) != freq.end()) {
freq[words[i]]++;
} else {
freq[words[i]] = 1;
}
}
int version = 0;
size_t i = 0;
size_t one_len = words[0].length();
size_t num = words.size();
size_t total_len = one_len * num;
while (i < s.length() - total_len + 1) {
i = findFirstOf(s, i, words);
if (i == string::npos) {
return ret;
}
string s1 = s.substr(i, one_len);
version++;
unordered_map<string, int> cnt;
cnt[s1] = 1;
//cout << "i=" << i << " version=" << version << " s1=" << s1 << endl;
int count = 0;
int k = i;
map[s1] = version;
//cout << "count=" << count << " k=" << k << " str=" << s1 << " it=" << version << endl;
k += one_len;
count++;
while (count < num) {
string s1 = s.substr(k, one_len);
auto it = map.find(s1);
//cout << "count=" << count << " k=" << k << " str=" << s1 << " it=" << (it == map.end() ? -1 : it->second) << endl;
if (it == map.end()) {
break;
}
if (it->second == version) {
cnt[s1]++;
if (cnt[s1] > freq[s1]) {
break;
}
} else {
cnt[s1] = 1;
}
map[s1] = version;
k += one_len;
count++;
}
if (count == num) {
ret.push_back(i);
//cout << "find idx: " << i << endl;
}
i++;
}
return ret;
}
size_t findFirstOf(string&s, size_t b, vector<string>& words) {
size_t min = s.length();
for (int i = 0; i < words.size(); i++) {
size_t cur = s.find(words[i], b);
//cout << "firstOf: " << words[i] << " cur=" << cur << " s=" << s.substr(b) << endl;
if (cur == string::npos) {
return string::npos;
}
if (cur < min) {
min = cur;
}
}
if (min == s.length()) {
return string::npos;
} else {
return min;
}
}
};
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