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feat: update lc problems #3697

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merged 1 commit into from
Oct 30, 2024
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feat: update lc problems #3697

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10 changes: 5 additions & 5 deletions lcof2/剑指 Offer II 109. 开密码锁/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -289,12 +289,12 @@ class Solution {
if target == "0000" {
return 0
}

var visited = Set<String>()
var queue = ["0000"]
visited.insert("0000")
var step = 0

while !queue.isEmpty {
step += 1
for _ in 0..<queue.count {
Expand All @@ -311,10 +311,10 @@ class Solution {
}
}
}

return -1
}

private func getNeighbors(_ lock: String) -> [String] {
var neighbors = [String]()
var chars = Array(lock)
Expand All @@ -328,7 +328,7 @@ class Solution {
}
return neighbors
}

private func prevChar(_ c: Character) -> Character {
return c == "0" ? "9" : Character(UnicodeScalar(c.asciiValue! - 1))
}
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2 changes: 1 addition & 1 deletion lcof2/剑指 Offer II 110. 所有路径/README.md
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Expand Up @@ -208,7 +208,7 @@ class Solution {
results.append(Array(path))
return
}

for next in graph[node] {
path.append(next)
dfs(next, &path)
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24 changes: 12 additions & 12 deletions lcof2/剑指 Offer II 111. 计算除法/README.md
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Expand Up @@ -298,23 +298,23 @@ func find(x int) int {
class Solution {
private var parent = [Int]()
private var weight = [Double]()

func calcEquation(
_ equations: [[String]],
_ values: [Double],
_ equations: [[String]],
_ values: [Double],
_ queries: [[String]]
) -> [Double] {
let n = equations.count
parent = Array(0..<(n * 2))
weight = Array(repeating: 1.0, count: n * 2)

var map = [String: Int]()
var index = 0

for i in 0..<n {
let a = equations[i][0]
let b = equations[i][1]

if map[a] == nil {
map[a] = index
index += 1
Expand All @@ -323,18 +323,18 @@ class Solution {
map[b] = index
index += 1
}

let pa = find(map[a]!)
let pb = find(map[b]!)

if pa != pb {
parent[pa] = pb
weight[pa] = weight[map[b]!] * values[i] / weight[map[a]!]
}
}

var result = [Double]()

for query in queries {
let (c, d) = (query[0], query[1])
if let id1 = map[c], let id2 = map[d], find(id1) == find(id2) {
Expand All @@ -343,10 +343,10 @@ class Solution {
result.append(-1.0)
}
}

return result
}

private func find(_ x: Int) -> Int {
if parent[x] != x {
let origin = parent[x]
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4 changes: 2 additions & 2 deletions lcof2/剑指 Offer II 112. 最长递增路径/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -219,7 +219,7 @@ class Solution {
m = matrix.count
n = matrix[0].count
memo = Array(repeating: Array(repeating: -1, count: n), count: m)

var ans = 0
for i in 0..<m {
for j in 0..<n {
Expand All @@ -235,7 +235,7 @@ class Solution {
}
var ans = 1
let dirs = [(-1, 0), (1, 0), (0, -1), (0, 1)]

for (dx, dy) in dirs {
let x = i + dx, y = j + dy
if x >= 0, x < m, y >= 0, y < n, matrix[x][y] > matrix[i][j] {
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10 changes: 5 additions & 5 deletions lcof2/剑指 Offer II 113. 课程顺序/README.md
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Expand Up @@ -276,34 +276,34 @@ class Solution {
func findOrder(_ numCourses: Int, _ prerequisites: [[Int]]) -> [Int] {
var graph = Array(repeating: [Int](), count: numCourses)
var indegree = Array(repeating: 0, count: numCourses)

for prereq in prerequisites {
let course = prereq[0]
let prereqCourse = prereq[1]
graph[prereqCourse].append(course)
indegree[course] += 1
}

var queue = [Int]()
for i in 0..<numCourses {
if indegree[i] == 0 {
queue.append(i)
}
}

var order = [Int]()
while !queue.isEmpty {
let course = queue.removeFirst()
order.append(course)

for nextCourse in graph[course] {
indegree[nextCourse] -= 1
if indegree[nextCourse] == 0 {
queue.append(nextCourse)
}
}
}

return order.count == numCourses ? order : []
}
}
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16 changes: 8 additions & 8 deletions lcof2/剑指 Offer II 114. 外星文字典/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -358,7 +358,7 @@ class Solution {
var indegree = Array(repeating: 0, count: 26)
var seen = Array(repeating: false, count: 26)
var letterCount = 0

for i in 0..<words.count - 1 {
for char in words[i] {
let index = Int(char.asciiValue! - Character("a").asciiValue!)
Expand All @@ -371,39 +371,39 @@ class Solution {
for j in 0..<minLength {
let char1 = words[i][words[i].index(words[i].startIndex, offsetBy: j)]
let char2 = words[i + 1][words[i + 1].index(words[i + 1].startIndex, offsetBy: j)]

if char1 != char2 {
let c1 = Int(char1.asciiValue! - Character("a").asciiValue!)
let c2 = Int(char2.asciiValue! - Character("a").asciiValue!)

if !graph[c1].contains(c2) {
graph[c1].insert(c2)
indegree[c2] += 1
}
break
}

if j == minLength - 1 && words[i].count > words[i + 1].count {
return ""
}
}
}

for char in words[words.count - 1] {
let index = Int(char.asciiValue! - Character("a").asciiValue!)
if !seen[index] {
seen[index] = true
letterCount += 1
}
}

var queue = [Int]()
for i in 0..<26 {
if seen[i] && indegree[i] == 0 {
queue.append(i)
}
}

var order = ""
while !queue.isEmpty {
let u = queue.removeFirst()
Expand All @@ -415,7 +415,7 @@ class Solution {
}
}
}

return order.count == letterCount ? order : ""
}
}
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112 changes: 58 additions & 54 deletions solution/2200-2299/2222.Number of Ways to Select Buildings/README.md
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Expand Up @@ -74,11 +74,13 @@ tags:

<!-- solution:start -->

### 方法一：统计 010 和 101 的出现次数
### 方法一：计数 + 枚举

有效方案只有两种情况：$010$ 和 $101$。枚举中间数字，累计方案数。
根据题目描述，我们需要选择 $3$ 栋建筑，且相邻的两栋不能是同一类型。

时间复杂度 $O(n)$，其中 $n$ 表示 $s$ 的长度。
我们可以枚举中间的建筑，假设为 $x$，那么左右两边的建筑类型只能是 $x \oplus 1$，其中 $\oplus$ 表示异或运算。因此，我们可以使用两个数组 $l$ 和 $r$ 分别记录左右两边的建筑类型的数量，然后枚举中间的建筑，计算答案即可。

时间复杂度 $O(n)$，其中 $n$ 是字符串 $s$ 的长度。空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand All @@ -87,18 +89,13 @@ tags:
```python
class Solution:
def numberOfWays(self, s: str) -> int:
n = len(s)
cnt0 = s.count("0")
cnt1 = n - cnt0
c0 = c1 = 0
l = [0, 0]
r = [s.count("0"), s.count("1")]
ans = 0
for c in s:
if c == "0":
ans += c1 * (cnt1 - c1)
c0 += 1
else:
ans += c0 * (cnt0 - c0)
c1 += 1
for x in map(int, s):
r[x] -= 1
ans += l[x ^ 1] * r[x ^ 1]
l[x] += 1
return ans
```

Expand All @@ -108,23 +105,17 @@ class Solution:
class Solution {
public long numberOfWays(String s) {
int n = s.length();
int cnt0 = 0;
for (char c : s.toCharArray()) {
if (c == '0') {
++cnt0;
}
int[] l = new int[2];
int[] r = new int[2];
for (int i = 0; i < n; ++i) {
r[s.charAt(i) - '0']++;
}
int cnt1 = n - cnt0;
long ans = 0;
int c0 = 0, c1 = 0;
for (char c : s.toCharArray()) {
if (c == '0') {
ans += c1 * (cnt1 - c1);
++c0;
} else {
ans += c0 * (cnt0 - c0);
++c1;
}
for (int i = 0; i < n; ++i) {
int x = s.charAt(i) - '0';
r[x]--;
ans += 1L * l[x ^ 1] * r[x ^ 1];
l[x]++;
}
return ans;
}
Expand All @@ -138,19 +129,16 @@ class Solution {
public:
long long numberOfWays(string s) {
int n = s.size();
int cnt0 = 0;
for (char& c : s) cnt0 += c == '0';
int cnt1 = n - cnt0;
int c0 = 0, c1 = 0;
int l[2]{};
int r[2]{};
r[0] = ranges::count(s, '0');
r[1] = n - r[0];
long long ans = 0;
for (char& c : s) {
if (c == '0') {
ans += c1 * (cnt1 - c1);
++c0;
} else {
ans += c0 * (cnt0 - c0);
++c1;
}
for (int i = 0; i < n; ++i) {
int x = s[i] - '0';
r[x]--;
ans += 1LL * l[x ^ 1] * r[x ^ 1];
l[x]++;
}
return ans;
}
Expand All @@ -160,22 +148,38 @@ public:
#### Go

```go
func numberOfWays(s string) int64 {
func numberOfWays(s string) (ans int64) {
n := len(s)
cnt0 := strings.Count(s, "0")
cnt1 := n - cnt0
c0, c1 := 0, 0
ans := 0
l := [2]int{}
r := [2]int{}
r[0] = strings.Count(s, "0")
r[1] = n - r[0]
for _, c := range s {
if c == '0' {
ans += c1 * (cnt1 - c1)
c0++
} else {
ans += c0 * (cnt0 - c0)
c1++
}
x := int(c - '0')
r[x]--
ans += int64(l[x^1] * r[x^1])
l[x]++
}
return int64(ans)
return
}
```

#### TypeScript

```ts
function numberOfWays(s: string): number {
const n = s.length;
const l: number[] = [0, 0];
const r: number[] = [s.split('').filter(c => c === '0').length, 0];
r[1] = n - r[0];
let ans: number = 0;
for (const c of s) {
const x = c === '0' ? 0 : 1;
r[x]--;
ans += l[x ^ 1] * r[x ^ 1];
l[x]++;
}
return ans;
}
```

Expand Down
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