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Commit dc95795

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‎note/030/README.md

Lines changed: 57 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -19,8 +19,8 @@ The output order does not matter, returning [9,0] is fine too.
1919

2020
```
2121
Input:
22-
s = "wordgoodstudentgoodword",
23-
words = ["word","student"]
22+
s = "wordgoodgoodgoodbestword",
23+
words = ["word","good","best","word"]
2424
Output: []
2525
```
2626

@@ -29,10 +29,63 @@ Output: []
2929

3030
## 思路
3131

32-
题意是
32+
题意是给自个字符串 `s` 和等长度的单词数组 `words`,我们要找到的就是在 `s` 串中由所有单词组成的子串的索引(不要求顺序),不明白的话看下例子也就理解了,比如例子 1 的结果 [0, 9]:[`barfoo`,`foobar`] 就是符合的子串。
3333

34-
```java
34+
我们把 `words` 每个单词出现的次数都存入到一个 `map` 中,然后遍历 `s` 串,依次截取单词长度的子串做比较,如果都符合那就加入结果,如果不符合,我们要把和它相关联的不符合的都剔除掉,这样在之后遍历我们就可以跳过了达到优化的目的。
3535

36+
```java
37+
public class Solution {
38+
public List<Integer> findSubstring(String s, String[] words) {
39+
if (s == null) return Collections.emptyList();
40+
int len = s.length();
41+
if (len == 0) return Collections.emptyList();
42+
int wordsSize = words.length;
43+
if (wordsSize == 0) return Collections.emptyList();
44+
int wordLen = words[0].length(), end = len - wordsSize * wordLen;
45+
if (end < 0) return Collections.emptyList();
46+
Map<String, Integer> countMap = new HashMap<>();
47+
for (String word : words) {
48+
countMap.put(word, countMap.getOrDefault(word, 0) + 1);
49+
}
50+
List<Integer> res = new ArrayList<>();
51+
Set<Integer> ignores = new HashSet<>();
52+
for (int i = 0; i <= end; ++i) {
53+
if (ignores.contains(i)) continue;
54+
Map<String, Integer> findMap = new HashMap<>();
55+
int st = i, count = 0;
56+
List<Integer> ignore = new ArrayList<>();
57+
for (int j = 0; ; ++j) {
58+
int cur = i + j * wordLen;
59+
if (cur + wordLen > len) break;
60+
String word = s.substring(cur, cur + wordLen);
61+
if (countMap.containsKey(word)) {
62+
findMap.put(word, findMap.getOrDefault(word, 0) + 1);
63+
++count;
64+
while (findMap.get(word) > countMap.get(word)) {
65+
ignore.add(st);
66+
String tmp = s.substring(st, st += wordLen);
67+
findMap.put(tmp, findMap.get(tmp) - 1);
68+
--count;
69+
}
70+
if (count == wordsSize) {
71+
ignore.add(st);
72+
res.add(st);
73+
String tmp = s.substring(st, st += wordLen);
74+
findMap.put(tmp, findMap.get(tmp) - 1);
75+
--count;
76+
}
77+
} else {
78+
for (int k = i; k <= cur; k += wordLen) {
79+
ignore.add(k);
80+
}
81+
break;
82+
}
83+
}
84+
ignores.addAll(ignore);
85+
}
86+
return res;
87+
}
88+
}
3689
```
3790

3891

‎note/031/README.md

Lines changed: 36 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,36 @@
1+
# [Next Permutation][title]
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3+
## Description
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5+
Implement **next permutation**, which rearranges numbers into the lexicographically next greater permutation of numbers.
6+
7+
If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order).
8+
9+
The replacement must be **in-place** and use only constant extra memory.
10+
11+
Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column.
12+
13+
`1,2,3` → `1,3,2`
14+
`3,2,1` → `1,2,3`
15+
`1,1,5` → `1,5,1`
16+
17+
**Tags:** Array
18+
19+
20+
## 思路
21+
22+
题意是
23+
24+
```java
25+
26+
```
27+
28+
29+
## 结语
30+
31+
如果你同我一样热爱数据结构、算法、LeetCode,可以关注我 GitHub 上的 LeetCode 题解:[awesome-java-leetcode][ajl]
32+
33+
34+
35+
[title]: https://leetcode.com/problems/next-permutation
36+
[ajl]: https://github.com/Blankj/awesome-java-leetcode

‎note/043/README.md

Lines changed: 3 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -46,12 +46,12 @@ class Solution {
4646
ans[i + j + 1] += c * (c2[j] - '0');
4747
}
4848
}
49-
for (int i = l - 1; i > 0; ++i) {
49+
for (int i = l - 1; i > 0; --i) {
5050
if (ans[i] > 9) {
5151
ans[i - 1] += ans[i] / 10;
52-
ans[i] %= 10;
52+
ans[i] %= 10;
5353
}
54-
}
54+
}
5555
StringBuilder sb = new StringBuilder();
5656
int i = 0;
5757
for (; ; ++i) if (ans[i] != 0) break;

‎src/com/blankj/hard/_030/Solution.java

Lines changed: 7 additions & 48 deletions
Original file line numberDiff line numberDiff line change
@@ -18,7 +18,12 @@
1818
*/
1919
public class Solution {
2020
public List<Integer> findSubstring(String s, String[] words) {
21-
int wordsSize = words.length, wordLen = words[0].length(), end = s.length() - wordsSize * wordLen;
21+
if (s == null) return Collections.emptyList();
22+
int len = s.length();
23+
if (len == 0) return Collections.emptyList();
24+
int wordsSize = words.length;
25+
if (wordsSize == 0) return Collections.emptyList();
26+
int wordLen = words[0].length(), end = len - wordsSize * wordLen;
2227
if (end < 0) return Collections.emptyList();
2328
Map<String, Integer> countMap = new HashMap<>();
2429
for (String word : words) {
@@ -33,7 +38,7 @@ public List<Integer> findSubstring(String s, String[] words) {
3338
List<Integer> ignore = new ArrayList<>();
3439
for (int j = 0; ; ++j) {
3540
int cur = i + j * wordLen;
36-
if (cur + wordLen > s.length()) break;
41+
if (cur + wordLen > len) break;
3742
String word = s.substring(cur, cur + wordLen);
3843
if (countMap.containsKey(word)) {
3944
findMap.put(word, findMap.getOrDefault(word, 0) + 1);
@@ -63,52 +68,6 @@ public List<Integer> findSubstring(String s, String[] words) {
6368
return res;
6469
}
6570

66-
// public List<Integer> findSubstring(String S, String[] L) {
67-
// List<Integer> res = new LinkedList<>();
68-
// int N = S.length();
69-
// int M = L.length; // *** length
70-
// int wl = L[0].length();
71-
// Map<String, Integer> map = new HashMap<>(), curMap = new HashMap<>();
72-
// for (String s : L) {
73-
// if (map.containsKey(s)) map.put(s, map.get(s) + 1);
74-
// else map.put(s, 1);
75-
// }
76-
// String str = null, tmp = null;
77-
// for (int i = 0; i < wl; i++) {
78-
// int count = 0; // remark: reset count
79-
// int start = i;
80-
// for (int r = i; r + wl <= N; r += wl) {
81-
// str = S.substring(r, r + wl);
82-
// if (map.containsKey(str)) {
83-
// if (curMap.containsKey(str)) curMap.put(str, curMap.get(str) + 1);
84-
// else curMap.put(str, 1);
85-
//
86-
// if (curMap.get(str) <= map.get(str)) count++;
87-
// if (count == M) {
88-
// res.add(start);
89-
// tmp = S.substring(start, start + wl);
90-
// curMap.put(tmp, curMap.get(tmp) - 1);
91-
// start += wl;
92-
// count--;
93-
// }
94-
// while (curMap.get(str) > map.get(str)) {
95-
// tmp = S.substring(start, start + wl);
96-
// curMap.put(tmp, curMap.get(tmp) - 1);
97-
// start += wl;
98-
// if (curMap.get(tmp) < map.get(tmp)) count--;
99-
//
100-
// }
101-
// } else {
102-
// curMap.clear();
103-
// count = 0;
104-
// start = r + wl;
105-
// }
106-
// }
107-
// curMap.clear();
108-
// }
109-
// return res;
110-
// }
111-
11271
public static void main(String[] args) {
11372
Solution solution = new Solution();
11473
System.out.println(solution.findSubstring("wordgoodgoodgoodbestword", new String[]{"word", "good", "best", "good"}));

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