Add solution 0609、0692、0890、1048、1442、1738

Author: halfrost

README.md

@@ -0,0 +1,23 @@
+package leetcode
+
+import "strings"
+
+func findDuplicate(paths []string) [][]string {
+	cache := make(map[string][]string)
+	for _, path := range paths {
+		parts := strings.Split(path, " ")
+		dir := parts[0]
+		for i := 1; i < len(parts); i++ {
+			bracketPosition := strings.IndexByte(parts[i], '(')
+			content := parts[i][bracketPosition+1 : len(parts[i])-1]
+			cache[content] = append(cache[content], dir+"/"+parts[i][:bracketPosition])
+		}
+	}
+	res := make([][]string, 0, len(cache))
+	for _, group := range cache {
+		if len(group) >= 2 {
+			res = append(res, group)
+		}
+	}
+	return res
+}

leetcode/0609.Find-Duplicate-File-in-System/609. Find Duplicate File in System.go

@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question609 struct {
+	para609
+	ans609
+}
+
+// para 是参数
+// one 代表第一个参数
+type para609 struct {
+	paths []string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans609 struct {
+	one [][]string
+}
+
+func Test_Problem609(t *testing.T) {
+
+	qs := []question609{
+
+		{
+			para609{[]string{"root/a 1.txt(abcd) 2.txt(efgh)", "root/c 3.txt(abcd)", "root/c/d 4.txt(efgh)", "root 4.txt(efgh)"}},
+			ans609{[][]string{{"root/a/2.txt", "root/c/d/4.txt", "root/4.txt"}, {"root/a/1.txt", "root/c/3.txt"}}},
+		},
+
+		{
+			para609{[]string{"root/a 1.txt(abcd) 2.txt(efgh)", "root/c 3.txt(abcd)", "root/c/d 4.txt(efgh)"}},
+			ans609{[][]string{{"root/a/2.txt", "root/c/d/4.txt"}, {"root/a/1.txt", "root/c/3.txt"}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 609------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans609, q.para609
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findDuplicate(p.paths))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0609.Find-Duplicate-File-in-System/609. Find Duplicate File in System_test.go

@@ -0,0 +1,93 @@
+# [609. Find Duplicate File in System](https://leetcode.com/problems/find-duplicate-file-in-system/)
+
+
+## 题目
+
+Given a list `paths` of directory info, including the directory path, and all the files with contents in this directory, return *all the duplicate files in the file system in terms of their paths*. You may return the answer in **any order**.
+
+A group of duplicate files consists of at least two files that have the same content.
+
+A single directory info string in the input list has the following format:
+
+- `"root/d1/d2/.../dm f1.txt(f1_content) f2.txt(f2_content) ... fn.txt(fn_content)"`
+
+It means there are `n` files `(f1.txt, f2.txt ... fn.txt)` with content `(f1_content, f2_content ... fn_content)` respectively in the directory "`root/d1/d2/.../dm"`. Note that `n >= 1` and `m >= 0`. If `m = 0`, it means the directory is just the root directory.
+
+The output is a list of groups of duplicate file paths. For each group, it contains all the file paths of the files that have the same content. A file path is a string that has the following format:
+
+- `"directory_path/file_name.txt"`
+
+**Example 1:**
+
+```
+Input: paths = ["root/a 1.txt(abcd) 2.txt(efgh)","root/c 3.txt(abcd)","root/c/d 4.txt(efgh)","root 4.txt(efgh)"]
+Output: [["root/a/2.txt","root/c/d/4.txt","root/4.txt"],["root/a/1.txt","root/c/3.txt"]]
+
+```
+
+**Example 2:**
+
+```
+Input: paths = ["root/a 1.txt(abcd) 2.txt(efgh)","root/c 3.txt(abcd)","root/c/d 4.txt(efgh)"]
+Output: [["root/a/2.txt","root/c/d/4.txt"],["root/a/1.txt","root/c/3.txt"]]
+
+```
+
+**Constraints:**
+
+- `1 <= paths.length <= 2 * 104`
+- `1 <= paths[i].length <= 3000`
+- `1 <= sum(paths[i].length) <= 5 * 105`
+- `paths[i]` consist of English letters, digits, `'/'`, `'.'`, `'('`, `')'`, and `' '`.
+- You may assume no files or directories share the same name in the same directory.
+- You may assume each given directory info represents a unique directory. A single blank space separates the directory path and file info.
+
+**Follow up:**
+
+- Imagine you are given a real file system, how will you search files? DFS or BFS?
+- If the file content is very large (GB level), how will you modify your solution?
+- If you can only read the file by 1kb each time, how will you modify your solution?
+- What is the time complexity of your modified solution? What is the most time-consuming part and memory-consuming part of it? How to optimize?
+- How to make sure the duplicated files you find are not false positive?
+
+## 题目大意
+
+给定一个目录信息列表，包括目录路径，以及该目录中的所有包含内容的文件，您需要找到文件系统中的所有重复文件组的路径。一组重复的文件至少包括二个具有完全相同内容的文件。输入列表中的单个目录信息字符串的格式如下：`"root/d1/d2/.../dm f1.txt(f1_content) f2.txt(f2_content) ... fn.txt(fn_content)"`。这意味着有 n 个文件（`f1.txt, f2.txt ... fn.txt` 的内容分别是 `f1_content, f2_content ... fn_content`）在目录 `root/d1/d2/.../dm` 下。注意：n>=1 且 m>=0。如果 m=0，则表示该目录是根目录。该输出是重复文件路径组的列表。对于每个组，它包含具有相同内容的文件的所有文件路径。文件路径是具有下列格式的字符串：`"directory_path/file_name.txt"`
+
+## 解题思路
+
+- 这一题算简单题，考察的是字符串基本操作与 map 的使用。首先通过字符串操作获取目录路径、文件名和文件内容。再使用 map 来寻找重复文件，key 是文件内容，value 是存储路径和文件名的列表。遍历每一个文件，并把它加入 map 中。最后遍历 map，如果一个键对应的值列表的长度大于 1，说明找到了重复文件，可以把这个列表加入到最终答案中。
+- 这道题有价值的地方在 **Follow up** 中。感兴趣的读者可以仔细想想以下几个问题：
+    1. 假设您有一个真正的文件系统，您将如何搜索文件？广度搜索还是宽度搜索？
+    2. 如果文件内容非常大（GB级别），您将如何修改您的解决方案？
+    3. 如果每次只能读取 1 kb 的文件，您将如何修改解决方案？
+    4. 修改后的解决方案的时间复杂度是多少？其中最耗时的部分和消耗内存的部分是什么？如何优化？
+    5. 如何确保您发现的重复文件不是误报？
+
+## 代码
+
+```go
+package leetcode
+
+import "strings"
+
+func findDuplicate(paths []string) [][]string {
+	cache := make(map[string][]string)
+	for _, path := range paths {
+		parts := strings.Split(path, " ")
+		dir := parts[0]
+		for i := 1; i < len(parts); i++ {
+			bracketPosition := strings.IndexByte(parts[i], '(')
+			content := parts[i][bracketPosition+1 : len(parts[i])-1]
+			cache[content] = append(cache[content], dir+"/"+parts[i][:bracketPosition])
+		}
+	}
+	res := make([][]string, 0, len(cache))
+	for _, group := range cache {
+		if len(group) >= 2 {
+			res = append(res, group)
+		}
+	}
+	return res
+}
+```

leetcode/0609.Find-Duplicate-File-in-System/README.md

@@ -0,0 +1,50 @@
+package leetcode
+
+import "container/heap"
+
+func topKFrequent(words []string, k int) []string {
+	m := map[string]int{}
+	for _, word := range words {
+		m[word]++
+	}
+	pq := &PQ{}
+	heap.Init(pq)
+	for w, c := range m {
+		heap.Push(pq, &wordCount{w, c})
+		if pq.Len() > k {
+			heap.Pop(pq)
+		}
+	}
+	res := make([]string, k)
+	for i := k - 1; i >= 0; i-- {
+		wc := heap.Pop(pq).(*wordCount)
+		res[i] = wc.word
+	}
+	return res
+}
+
+type wordCount struct {
+	word string
+	cnt  int
+}
+
+type PQ []*wordCount
+
+func (pq PQ) Len() int      { return len(pq) }
+func (pq PQ) Swap(i, j int) { pq[i], pq[j] = pq[j], pq[i] }
+func (pq PQ) Less(i, j int) bool {
+	if pq[i].cnt == pq[j].cnt {
+		return pq[i].word > pq[j].word
+	}
+	return pq[i].cnt < pq[j].cnt
+}
+func (pq *PQ) Push(x interface{}) {
+	tmp := x.(*wordCount)
+	*pq = append(*pq, tmp)
+}
+func (pq *PQ) Pop() interface{} {
+	n := len(*pq)
+	tmp := (*pq)[n-1]
+	*pq = (*pq)[:n-1]
+	return tmp
+}

leetcode/0692.Top-K-Frequent-Words/692. Top K Frequent Words.go

@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question692 struct {
+	para692
+	ans692
+}
+
+// para 是参数
+// one 代表第一个参数
+type para692 struct {
+	words []string
+	k     int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans692 struct {
+	one []string
+}
+
+func Test_Problem692(t *testing.T) {
+
+	qs := []question692{
+
+		{
+			para692{[]string{"i", "love", "leetcode", "i", "love", "coding"}, 2},
+			ans692{[]string{"i", "love"}},
+		},
+
+		{
+			para692{[]string{"the", "day", "is", "sunny", "the", "the", "the", "sunny", "is", "is"}, 4},
+			ans692{[]string{"the", "is", "sunny", "day"}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 692------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans692, q.para692
+		fmt.Printf("【input】:%v       【output】:%v\n", p, topKFrequent(p.words, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0692.Top-K-Frequent-Words/692. Top K Frequent Words_test.go

@@ -0,0 +1,98 @@
+# [692. Top K Frequent Words](https://leetcode.com/problems/top-k-frequent-words/)
+
+
+## 题目
+
+Given a non-empty list of words, return the k most frequent elements.
+
+Your answer should be sorted by frequency from highest to lowest. If two words have the same frequency, then the word with the lower alphabetical order comes first.
+
+**Example 1:**
+
+```
+Input: ["i", "love", "leetcode", "i", "love", "coding"], k = 2
+Output: ["i", "love"]
+Explanation: "i" and "love" are the two most frequent words.
+    Note that "i" comes before "love" due to a lower alphabetical order.
+```
+
+**Example 2:**
+
+```
+Input: ["the", "day", "is", "sunny", "the", "the", "the", "sunny", "is", "is"], k = 4
+Output: ["the", "is", "sunny", "day"]
+Explanation: "the", "is", "sunny" and "day" are the four most frequent words,
+    with the number of occurrence being 4, 3, 2 and 1 respectively.
+```
+
+**Note:**
+
+1. You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
+2. Input words contain only lowercase letters.
+
+**Follow up:**
+
+1. Try to solve it in O(n log k) time and O(n) extra space.
+
+## 题目大意
+
+给一非空的单词列表，返回前 *k* 个出现次数最多的单词。返回的答案应该按单词出现频率由高到低排序。如果不同的单词有相同出现频率，按字母顺序排序。
+
+## 解题思路
+
+- 思路很简单的题。维护一个长度为 k 的最大堆，先按照频率排，如果频率相同再按照字母顺序排。最后输出依次将优先队列里面的元素 pop 出来即可。
+
+## 代码
+
+```go
+package leetcode
+
+import "container/heap"
+
+func topKFrequent(words []string, k int) []string {
+	m := map[string]int{}
+	for _, word := range words {
+		m[word]++
+	}
+	pq := &PQ{}
+	heap.Init(pq)
+	for w, c := range m {
+		heap.Push(pq, &wordCount{w, c})
+		if pq.Len() > k {
+			heap.Pop(pq)
+		}
+	}
+	res := make([]string, k)
+	for i := k - 1; i >= 0; i-- {
+		wc := heap.Pop(pq).(*wordCount)
+		res[i] = wc.word
+	}
+	return res
+}
+
+type wordCount struct {
+	word string
+	cnt  int
+}
+
+type PQ []*wordCount
+
+func (pq PQ) Len() int      { return len(pq) }
+func (pq PQ) Swap(i, j int) { pq[i], pq[j] = pq[j], pq[i] }
+func (pq PQ) Less(i, j int) bool {
+	if pq[i].cnt == pq[j].cnt {
+		return pq[i].word > pq[j].word
+	}
+	return pq[i].cnt < pq[j].cnt
+}
+func (pq *PQ) Push(x interface{}) {
+	tmp := x.(*wordCount)
+	*pq = append(*pq, tmp)
+}
+func (pq *PQ) Pop() interface{} {
+	n := len(*pq)
+	tmp := (*pq)[n-1]
+	*pq = (*pq)[:n-1]
+	return tmp
+}
+```

leetcode/0692.Top-K-Frequent-Words/README.md

@@ -0,0 +1,32 @@
+package leetcode
+
+func findAndReplacePattern(words []string, pattern string) []string {
+	res := make([]string, 0)
+	for _, word := range words {
+		if match(word, pattern) {
+			res = append(res, word)
+		}
+	}
+	return res
+}
+
+func match(w, p string) bool {
+	if len(w) != len(p) {
+		return false
+	}
+	m, used := make(map[uint8]uint8), make(map[uint8]bool)
+	for i := 0; i < len(w); i++ {
+		if v, ok := m[p[i]]; ok {
+			if w[i] != v {
+				return false
+			}
+		} else {
+			if used[w[i]] {
+				return false
+			}
+			m[p[i]] = w[i]
+			used[w[i]] = true
+		}
+	}
+	return true
+}