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[LeetCode] 30. 串联所有单词的子串
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题目链接: https://leetcode-cn.com/problems/substring-with-concatenation-of-all-words/
题目描述:
给定一个字符串 s 和一些长度相同的单词 words。 找出 s 中恰好可以由 words 中所有单词串联形成的子串的起始位置。
注意子串要与 words 中的单词完全匹配，中间不能有其他字符，但不需要考虑 words 中单词串联的顺序。
示例:
示例 1：
输入：
s = "barfoothefoobarman",
words = ["foo","bar"]
输出：[0,9]
解释：
从索引 0 和 9 开始的子串分别是 "barfoor" 和 "foobar" 。
输出的顺序不重要, [9,0] 也是有效答案。
示例 2：
输入：
s = "wordgoodgoodgoodbestword",
words = ["word","good","best","word"]
输出：[]
思路:
一开始,我的想法是,每次从
s
截取一定长度(固定的)的字符串,看这段字符串出现单词个数是否和要匹配的单词个数相等!如下代码:
class Solution:
def findSubstring(self, s: str, words: List[str]) -> List[int]:
from collections import Counter
if not s or not words:return []
all_len = sum(map(len, words))
n = len(s)
words = Counter(words)
res = []
for i in range(0, n - all_len + 1):
tmp = s[i:i+all_len]
flag = True
for key in words:
if words[key] != tmp.count(key):
flag = False
break
if flag:res.append(i)
return res
但是比如:
s = "ababaab" ,words = ["ab","ba","ba"]
就会报错!
错误原因,因为计算时候我们会从字符串中间计算,也就是说会出现单词截断的问题.
所以我想另一种方法,
思路一:
因为单词长度固定的,我们可以计算出截取字符串的单词个数是否和
words
里相等,所以我们可以借用哈希表.
一个是哈希表是
words
,一个哈希表是截取的字符串,
比较两个哈希是否相等!
因为遍历和比较都是线性的,所以时间复杂度 :\(O(n^2)\)
上面思路每次都要反复遍历
s
;
下面介绍滑动窗口.
思路二:
滑动窗口!
我们一直在
s
维护着所有单词长度总和的一个长度队列!
关于滑动窗口,可以看看这篇文章.
时间复杂度:\(O(n)\)
还可以再优化,只是加一些判断,详细看代码吧!
关注我的知乎专栏,了解更多的解题技巧,大家一起进步,加油!
代码:
思路一:
python
class Solution:
def findSubstring(self, s: str, words: List[str]) -> List[int]:
from collections import Counter
if not s or not words:return []
one_word = len(words[0])
all_len = len(words) * one_word
n = len(s)
words = Counter(words)
res = []
for i in range(0, n - all_len + 1):
tmp = s[i:i+all_len]
c_tmp = []
for j in range(0, all_len, one_word):
c_tmp.append(tmp[j:j+one_word])
if Counter(c_tmp) == words:
res.append(i)
return res
java
class Solution {
public List<Integer> findSubstring(String s, String[] words) {
List<Integer> res = new ArrayList<>();
if (s == null || s.length() == 0 || words == null || words.length == 0) return res;
HashMap<String, Integer> map = new HashMap<>();
int one_word = words[0].length();
int word_num = words.length;
int all_len = one_word * word_num;
for (String word : words) {
map.put(word, map.getOrDefault(word, 0) + 1);
}
for (int i = 0; i < s.length() - all_len + 1; i++) {
String tmp = s.substring(i, i + all_len);
HashMap<String, Integer> tmp_map = new HashMap<>();
for (int j = 0; j < all_len; j += one_word) {
String w = tmp.substring(j, j + one_word);
tmp_map.put(w, tmp_map.getOrDefault(w, 0) + 1);
}
if (map.equals(tmp_map)) res.add(i);
}
return res;
}
}
思路二:
python
class Solution:
def findSubstring(self, s: str, words: List[str]) -> List[int]:
from collections import Counter
if not s or not words:return []
one_word = len(words[0])
word_num = len(words)
n = len(s)
words = Counter(words)
res = []
for i in range(0, one_word):
cur_cnt = 0
left = i
right = i
cur_Counter = Counter()
while right + one_word <= n:
w = s[right:right + one_word]
right += one_word
cur_Counter[w] += 1
cur_cnt += 1
while cur_Counter[w] > words[w]:
left_w = s[left:left+one_word]
left += one_word
cur_Counter[left_w] -= 1
cur_cnt -= 1
if cur_cnt == word_num :
res.append(left)
return res
java
class Solution {
public List<Integer> findSubstring(String s, String[] words) {
List<Integer> res = new ArrayList<>();
if (s == null || s.length() == 0 || words == null || words.length == 0) return res;
HashMap<String, Integer> map = new HashMap<>();
int one_word = words[0].length();
int word_num = words.length;
int all_len = one_word * word_num;
for (String word : words) {
map.put(word, map.getOrDefault(word, 0) + 1);
}
for (int i = 0; i < one_word; i++) {
int left = i, right = i, count = 0;
HashMap<String, Integer> tmp_map = new HashMap<>();
while (right + one_word <= s.length()) {
String w = s.substring(right, right + one_word);
tmp_map.put(w, tmp_map.getOrDefault(w, 0) + 1);
right += one_word;
count++;
while (tmp_map.getOrDefault(w, 0) > map.getOrDefault(w, 0)) {
String t_w = s.substring(left, left + one_word);
count--;
tmp_map.put(t_w, tmp_map.getOrDefault(t_w, 0) - 1);
left += one_word;
}
if (count == word_num) res.add(left);
}
}
return res;
}
}
再优化:
python
class Solution:
def findSubstring(self, s: str, words: List[str]) -> List[int]:
from collections import Counter
if not s or not words:return []
one_word = len(words[0])
word_num = len(words)
n = len(s)
if n < one_word:return []
words = Counter(words)
res = []
for i in range(0, one_word):
cur_cnt = 0
left = i
right = i
cur_Counter = Counter()
while right + one_word <= n:
w = s[right:right + one_word]
right += one_word
if w not in words:
left = right
cur_Counter.clear()
cur_cnt = 0
else:
cur_Counter[w] += 1
cur_cnt += 1
while cur_Counter[w] > words[w]:
left_w = s[left:left+one_word]
left += one_word
cur_Counter[left_w] -= 1
cur_cnt -= 1
if cur_cnt == word_num :
res.append(left)
return res
java
class Solution {
public List<Integer> findSubstring(String s, String[] words) {
List<Integer> res = new ArrayList<>();
if (s == null || s.length() == 0 || words == null || words.length == 0) return res;
HashMap<String, Integer> map = new HashMap<>();
int one_word = words[0].length();
int word_num = words.length;
int all_len = one_word * word_num;
for (String word : words) {
map.put(word, map.getOrDefault(word, 0) + 1);
}
for (int i = 0; i < one_word; i++) {
int left = i, right = i, count = 0;
HashMap<String, Integer> tmp_map = new HashMap<>();
while (right + one_word <= s.length()) {
String w = s.substring(right, right + one_word);
right += one_word;
if (!map.containsKey(w)) {
count = 0;
left = right;
tmp_map.clear();
} else {
tmp_map.put(w, tmp_map.getOrDefault(w, 0) + 1);
count++;
while (tmp_map.getOrDefault(w, 0) > map.getOrDefault(w, 0)) {
String t_w = s.substring(left, left + one_word);
count--;
tmp_map.put(t_w, tmp_map.getOrDefault(t_w, 0) - 1);
left += one_word;
}
if (count == word_num) res.add(left);
}
}
}
return res;
}
}
