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Blankj
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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
```
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Input:
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s = "wordgoodstudentgoodword",
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words = ["word","student"]
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s = "wordgoodgoodgoodbestword",
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words = ["word","good","best","word"]
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Output: []
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```
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@@ -29,10 +29,63 @@ Output: []
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## 思路
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题意是
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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。
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```java
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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。
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```java
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public class Solution {
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public List<Integer> findSubstring(String s, String[] words) {
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if (s == null) return Collections.emptyList();
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int len = s.length();
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if (len == 0) return Collections.emptyList();
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int wordsSize = words.length;
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if (wordsSize == 0) return Collections.emptyList();
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int wordLen = words[0].length(), end = len - wordsSize * wordLen;
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if (end < 0) return Collections.emptyList();
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Map<String, Integer> countMap = new HashMap<>();
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for (String word : words) {
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countMap.put(word, countMap.getOrDefault(word, 0) + 1);
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}
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List<Integer> res = new ArrayList<>();
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Set<Integer> ignores = new HashSet<>();
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for (int i = 0; i <= end; ++i) {
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if (ignores.contains(i)) continue;
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Map<String, Integer> findMap = new HashMap<>();
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int st = i, count = 0;
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List<Integer> ignore = new ArrayList<>();
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for (int j = 0; ; ++j) {
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int cur = i + j * wordLen;
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if (cur + wordLen > len) break;
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String word = s.substring(cur, cur + wordLen);
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if (countMap.containsKey(word)) {
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findMap.put(word, findMap.getOrDefault(word, 0) + 1);
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++count;
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while (findMap.get(word) > countMap.get(word)) {
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ignore.add(st);
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String tmp = s.substring(st, st += wordLen);
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findMap.put(tmp, findMap.get(tmp) - 1);
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--count;
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}
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if (count == wordsSize) {
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ignore.add(st);
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res.add(st);
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String tmp = s.substring(st, st += wordLen);
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findMap.put(tmp, findMap.get(tmp) - 1);
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--count;
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}
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} else {
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for (int k = i; k <= cur; k += wordLen) {
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ignore.add(k);
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}
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break;
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}
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}
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ignores.addAll(ignore);
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}
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return res;
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}
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}
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```
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‎note/031/README.md‎

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,36 @@
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# [Next Permutation][title]
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## Description
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Implement **next permutation**, which rearranges numbers into the lexicographically next greater permutation of numbers.
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If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order).
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The replacement must be **in-place** and use only constant extra memory.
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Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column.
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`1,2,3` → `1,3,2`
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`3,2,1` → `1,2,3`
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`1,1,5` → `1,5,1`
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**Tags:** Array
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## 思路
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题意是
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```java
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```
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## 结语
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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]
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[title]: https://leetcode.com/problems/next-permutation
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[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 {
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ans[i + j + 1] += c * (c2[j] - '0');
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}
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}
49-
for (int i = l - 1; i > 0; ++i) {
49+
for (int i = l - 1; i > 0; --i) {
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if (ans[i] > 9) {
5151
ans[i - 1] += ans[i] / 10;
52-
ans[i] %= 10;
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ans[i] %= 10;
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}
54-
}
54+
}
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StringBuilder sb = new StringBuilder();
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int i = 0;
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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 @@
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*/
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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();
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if (len == 0) return Collections.emptyList();
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int wordsSize = words.length;
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if (wordsSize == 0) return Collections.emptyList();
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int wordLen = words[0].length(), end = len - wordsSize * wordLen;
2227
if (end < 0) return Collections.emptyList();
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Map<String, Integer> countMap = new HashMap<>();
2429
for (String word : words) {
@@ -33,7 +38,7 @@ public List<Integer> findSubstring(String s, String[] words) {
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List<Integer> ignore = new ArrayList<>();
3439
for (int j = 0; ; ++j) {
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int cur = i + j * wordLen;
36-
if (cur + wordLen > s.length()) break;
41+
if (cur + wordLen > len) break;
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String word = s.substring(cur, cur + wordLen);
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if (countMap.containsKey(word)) {
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findMap.put(word, findMap.getOrDefault(word, 0) + 1);
@@ -63,52 +68,6 @@ public List<Integer> findSubstring(String s, String[] words) {
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return res;
6469
}
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66-
// public List<Integer> findSubstring(String S, String[] L) {
67-
// List<Integer> res = new LinkedList<>();
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// int N = S.length();
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// int M = L.length; // *** length
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// int wl = L[0].length();
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// Map<String, Integer> map = new HashMap<>(), curMap = new HashMap<>();
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// for (String s : L) {
73-
// if (map.containsKey(s)) map.put(s, map.get(s) + 1);
74-
// else map.put(s, 1);
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// }
76-
// String str = null, tmp = null;
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// for (int i = 0; i < wl; i++) {
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// int count = 0; // remark: reset count
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// 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);
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// tmp = S.substring(start, start + wl);
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// curMap.put(tmp, curMap.get(tmp) - 1);
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// start += wl;
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// 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);
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// start += wl;
98-
// if (curMap.get(tmp) < map.get(tmp)) count--;
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//
100-
// }
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// } else {
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// curMap.clear();
103-
// count = 0;
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// start = r + wl;
105-
// }
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// }
107-
// curMap.clear();
108-
// }
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// return res;
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// }
111-
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public static void main(String[] args) {
11372
Solution solution = new Solution();
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System.out.println(solution.findSubstring("wordgoodgoodgoodbestword", new String[]{"word", "good", "best", "good"}));

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