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feat: add solutions to lc problem: No.2103 #1914

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Nov 2, 2023
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feat: add solutions to lc problem: No.2103 #1914

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feat: add solutions to lc problem: No.2103
yanglbme Nov 2, 2023
c96a3b6
fix: c code
yanglbme Nov 2, 2023
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138 changes: 85 additions & 53 deletions solution/2100-2199/2103.Rings and Rods/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -68,7 +68,15 @@

<!-- 这里可写通用的实现逻辑 -->

哈希表实现。
**方法一：位运算**

我们可以用一个长度为 $10$ 的数组 $mask$ 来表示每根杆上的环的颜色情况，其中 $mask[i]$ 表示第 $i$ 根杆上的环的颜色情况，如果第 $i$ 根杆上有红色、绿色、蓝色的环，那么 $mask[i]$ 的二进制表示为 $111$，即 $mask[i] = 7$。

我们遍历字符串 $rings$，对于每个颜色位置对 $(c, j)$，其中 $c$ 表示环的颜色，$j$ 表示环所在的杆的编号，我们将 $mask[j]$ 对应的二进制位进行置位，即 $mask[j] |= d[c]$，其中 $d[c]$ 表示颜色 $c$ 对应的二进制位。

最后我们统计 $mask$ 中值为 $7$ 的元素的个数，即为集齐全部三种颜色环的杆的数目。

时间复杂度 $O(n)$，空间复杂度 $O(|\Sigma|)$，其中 $n$ 表示字符串 $rings$ 的长度，而 $|\Sigma|$ 表示字符集的大小。

<!-- tabs:start -->

Expand All @@ -79,11 +87,13 @@
```python
class Solution:
def countPoints(self, rings: str) -> int:
mp = defaultdict(set)
for i in range(1, len(rings), 2):
c = int(rings[i])
mp[c].add(rings[i - 1])
return sum(len(v) == 3 for v in mp.values())
mask = [0] * 10
d = {"R": 1, "G": 2, "B": 4}
for i in range(0, len(rings), 2):
c = rings[i]
j = int(rings[i + 1])
mask[j] |= d[c]
return mask.count(7)
```

### **Java**
Expand All @@ -93,14 +103,19 @@ class Solution:
```java
class Solution {
public int countPoints(String rings) {
Map<Integer, Set<Character>> mp = new HashMap<>();
for (int i = 1; i < rings.length(); i += 2) {
int c = rings.charAt(i) - '0';
mp.computeIfAbsent(c, k -> new HashSet<>()).add(rings.charAt(i - 1));
int[] d = new int['Z'];
d['R'] = 1;
d['G'] = 2;
d['B'] = 4;
int[] mask = new int[10];
for (int i = 0, n = rings.length(); i < n; i += 2) {
int c = rings.charAt(i);
int j = rings.charAt(i + 1) - '0';
mask[j] |= d[c];
}
int ans = 0;
for (Set<Character> e : mp.values()) {
if (e.size() == 3) {
for (int x : mask) {
if (x == 7) {
++ans;
}
}
Expand All @@ -115,54 +130,54 @@ class Solution {
class Solution {
public:
int countPoints(string rings) {
unordered_map<int, unordered_set<char>> mp;
for (int i = 1; i < rings.size(); i += 2) {
int c = rings[i] - '0';
mp[c].insert(rings[i - 1]);
int d['Z']{['R'] = 1, ['G'] = 2, ['B'] = 4};
int mask[10]{};
for (int i = 0, n = rings.size(); i < n; i += 2) {
int c = rings[i];
int j = rings[i + 1] - '0';
mask[j] |= d[c];
}
int ans = 0;
for (int i = 0; i < 10; ++i)
if (mp[i].size() == 3)
++ans;
return ans;
return count(mask, mask + 10, 7);
}
};
```

### **Go**

```go
func countPoints(rings string) int {
mp := make(map[byte]map[byte]bool)
for i := 1; i < len(rings); i += 2 {
func countPoints(rings string) (ans int) {
d := ['Z']int{'R': 1, 'G': 2, 'B': 4}
mask := [10]int{}
for i, n := 0, len(rings); i < n; i += 2 {
c := rings[i]
if len(mp[c]) == 0 {
mp[c] = make(map[byte]bool)
}
mp[c][rings[i-1]] = true
j := int(rings[i+1] - '0')
mask[j] |= d[c]
}
ans := 0
for _, v := range mp {
if len(v) == 3 {
for _, x := range mask {
if x == 7 {
ans++
}
}
return ans
return
}
```

### **TypeScript**

```ts
function countPoints(rings: string): number {
const helper = (c: string) => c.charCodeAt(0) - 'A'.charCodeAt(0);
const n = rings.length;
const target = (1 << helper('R')) + (1 << helper('G')) + (1 << helper('B'));
const count = new Array(10).fill(0);
for (let i = 0; i < n; i += 2) {
count[rings[i + 1]] |= 1 << helper(rings[i]);
const idx = (c: string) => c.charCodeAt(0) - 'A'.charCodeAt(0);
const d: number[] = Array(26).fill(0);
d[idx('R')] = 1;
d[idx('G')] = 2;
d[idx('B')] = 4;
const mask: number[] = Array(10).fill(0);
for (let i = 0; i < rings.length; i += 2) {
const c = rings[i];
const j = rings[i + 1].charCodeAt(0) - '0'.charCodeAt(0);
mask[j] |= d[idx(c)];
}
return count.reduce((r, v) => (r += v === target ? 1 : 0), 0);
return mask.filter(x => x === 7).length;
}
```

Expand All @@ -171,16 +186,22 @@ function countPoints(rings: string): number {
```rust
impl Solution {
pub fn count_points(rings: String) -> i32 {
let rings = rings.as_bytes();
let target = (1 << b'R' - b'A') + (1 << b'G' - b'A') + (1 << b'B' - b'A');
let n = rings.len();
let mut count = [0; 10];
let mut i = 0;
while i < n {
count[(rings[i + 1] - b'0') as usize] |= 1 << rings[i] - b'A';
i += 2;
let mut d: [i32; 90] = [0; 90];
d['R' as usize] = 1;
d['G' as usize] = 2;
d['B' as usize] = 4;

let mut mask: [i32; 10] = [0; 10];

let cs: Vec<char> = rings.chars().collect();

for i in (0..cs.len()).step_by(2) {
let c = cs[i] as usize;
let j = cs[i + 1] as usize - '0' as usize;
mask[j] |= d[c];
}
count.iter().filter(|&v| *v == target).count() as i32

mask.iter().filter(|&&x| x == 7).count() as i32
}
}
```
Expand All @@ -189,17 +210,28 @@ impl Solution {

```c
int countPoints(char* rings) {
int target = (1 << ('R' - 'A')) + (1 << ('G' - 'A')) + (1 << ('B' - 'A'));
int count[10] = {0};
for (int i = 0; rings[i]; i += 2) {
count[rings[i + 1] - '0'] |= 1 << (rings[i] - 'A');
int d['Z'];
memset(d, 0, sizeof(d));
d['R'] = 1;
d['G'] = 2;
d['B'] = 4;

int mask[10];
memset(mask, 0, sizeof(mask));

for (int i = 0, n = strlen(rings); i < n; i += 2) {
int c = rings[i];
int j = rings[i + 1] - '0';
mask[j] |= d[c];
}

int ans = 0;
for (int i = 0; i < 10; i++) {
if (count[i] == target) {
if (mask[i] == 7) {
ans++;
}
}

return ans;
}
```
Expand Down
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