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Commit 6e8cd0d

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feat: update solutions to lc problems: No.2278,2279 (doocs#3377)
* No.2278.Percentage of Letter in String * No.2279.Maximum Bags With Full Capacity of Rocks
1 parent 51b11dc commit 6e8cd0d

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14 files changed

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-257
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14 files changed

+204
-257
lines changed

‎solution/2200-2299/2278.Percentage of Letter in String/README.md‎

Lines changed: 11 additions & 24 deletions
Original file line numberDiff line numberDiff line change
@@ -55,7 +55,11 @@ tags:
5555

5656
<!-- solution:start -->
5757

58-
### 方法一
58+
### 方法一:计数
59+
60+
我们可以遍历字符串 $\textit{s},ドル统计其中等于 $\textit{letter}$ 的字符的个数,然后根据公式 $\textit{count} \times 100 ,円 / ,円 \textit{len}(\textit{s})$ 计算百分比。
61+
62+
时间复杂度 $O(n),ドル其中 $n$ 为字符串 $\textit{s}$ 的长度。空间复杂度 $O(1)$。
5963

6064
<!-- tabs:start -->
6165

@@ -89,9 +93,7 @@ class Solution {
8993
class Solution {
9094
public:
9195
int percentageLetter(string s, char letter) {
92-
int cnt = 0;
93-
for (char& c : s) cnt += c == letter;
94-
return cnt * 100 / s.size();
96+
return 100 * ranges::count(s, letter) / s.size();
9597
}
9698
};
9799
```
@@ -100,26 +102,16 @@ public:
100102
101103
```go
102104
func percentageLetter(s string, letter byte) int {
103-
cnt := 0
104-
for i := range s {
105-
if s[i] == letter {
106-
cnt++
107-
}
108-
}
109-
return cnt * 100 / len(s)
105+
return strings.Count(s, string(letter)) * 100 / len(s)
110106
}
111107
```
112108

113109
#### TypeScript
114110

115111
```ts
116112
function percentageLetter(s: string, letter: string): number {
117-
let count = 0;
118-
let total = s.length;
119-
for (let i of s) {
120-
if (i === letter) count++;
121-
}
122-
return Math.floor((count / total) * 100);
113+
const count = s.split('').filter(c => c === letter).length;
114+
return Math.floor((100 * count) / s.length);
123115
}
124116
```
125117

@@ -128,13 +120,8 @@ function percentageLetter(s: string, letter: string): number {
128120
```rust
129121
impl Solution {
130122
pub fn percentage_letter(s: String, letter: char) -> i32 {
131-
let mut count = 0;
132-
for c in s.chars() {
133-
if c == letter {
134-
count += 1;
135-
}
136-
}
137-
((count * 100) / s.len()) as i32
123+
let count = s.chars().filter(|&c| c == letter).count();
124+
(100 * count as i32 / s.len() as i32) as i32
138125
}
139126
}
140127
```

‎solution/2200-2299/2278.Percentage of Letter in String/README_EN.md‎

Lines changed: 11 additions & 24 deletions
Original file line numberDiff line numberDiff line change
@@ -53,7 +53,11 @@ The percentage of characters in s that equal the letter &#39;k&#39; is 0%, so we
5353

5454
<!-- solution:start -->
5555

56-
### Solution 1
56+
### Solution 1: Counting
57+
58+
We can traverse the string $\textit{s}$ and count the number of characters that are equal to $\textit{letter}$. Then, we calculate the percentage using the formula $\textit{count} \times 100 ,円 / ,円 \textit{len}(\textit{s})$.
59+
60+
Time complexity is $O(n),ドル where $n$ is the length of the string $\textit{s}$. Space complexity is $O(1)$.
5761

5862
<!-- tabs:start -->
5963

@@ -87,9 +91,7 @@ class Solution {
8791
class Solution {
8892
public:
8993
int percentageLetter(string s, char letter) {
90-
int cnt = 0;
91-
for (char& c : s) cnt += c == letter;
92-
return cnt * 100 / s.size();
94+
return 100 * ranges::count(s, letter) / s.size();
9395
}
9496
};
9597
```
@@ -98,26 +100,16 @@ public:
98100
99101
```go
100102
func percentageLetter(s string, letter byte) int {
101-
cnt := 0
102-
for i := range s {
103-
if s[i] == letter {
104-
cnt++
105-
}
106-
}
107-
return cnt * 100 / len(s)
103+
return strings.Count(s, string(letter)) * 100 / len(s)
108104
}
109105
```
110106

111107
#### TypeScript
112108

113109
```ts
114110
function percentageLetter(s: string, letter: string): number {
115-
let count = 0;
116-
let total = s.length;
117-
for (let i of s) {
118-
if (i === letter) count++;
119-
}
120-
return Math.floor((count / total) * 100);
111+
const count = s.split('').filter(c => c === letter).length;
112+
return Math.floor((100 * count) / s.length);
121113
}
122114
```
123115

@@ -126,13 +118,8 @@ function percentageLetter(s: string, letter: string): number {
126118
```rust
127119
impl Solution {
128120
pub fn percentage_letter(s: String, letter: char) -> i32 {
129-
let mut count = 0;
130-
for c in s.chars() {
131-
if c == letter {
132-
count += 1;
133-
}
134-
}
135-
((count * 100) / s.len()) as i32
121+
let count = s.chars().filter(|&c| c == letter).count();
122+
(100 * count as i32 / s.len() as i32) as i32
136123
}
137124
}
138125
```
Lines changed: 2 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -1,8 +1,6 @@
11
class Solution {
22
public:
33
int percentageLetter(string s, char letter) {
4-
int cnt = 0;
5-
for (char& c : s) cnt += c == letter;
6-
return cnt * 100 / s.size();
4+
return 100 * ranges::count(s, letter) / s.size();
75
}
8-
};
6+
};
Lines changed: 2 additions & 8 deletions
Original file line numberDiff line numberDiff line change
@@ -1,9 +1,3 @@
11
func percentageLetter(s string, letter byte) int {
2-
cnt := 0
3-
for i := range s {
4-
if s[i] == letter {
5-
cnt++
6-
}
7-
}
8-
return cnt * 100 / len(s)
9-
}
2+
return strings.Count(s, string(letter)) * 100 / len(s)
3+
}
Lines changed: 2 additions & 7 deletions
Original file line numberDiff line numberDiff line change
@@ -1,11 +1,6 @@
11
impl Solution {
22
pub fn percentage_letter(s: String, letter: char) -> i32 {
3-
let mut count = 0;
4-
for c in s.chars() {
5-
if c == letter {
6-
count += 1;
7-
}
8-
}
9-
((count * 100) / s.len()) as i32
3+
let count = s.chars().filter(|&c| c == letter).count();
4+
(100 * count as i32 / s.len() as i32) as i32
105
}
116
}
Lines changed: 2 additions & 6 deletions
Original file line numberDiff line numberDiff line change
@@ -1,8 +1,4 @@
11
function percentageLetter(s: string, letter: string): number {
2-
let count = 0;
3-
let total = s.length;
4-
for (let i of s) {
5-
if (i === letter) count++;
6-
}
7-
return Math.floor((count / total) * 100);
2+
const count = s.split('').filter(c => c === letter).length;
3+
return Math.floor((100 * count) / s.length);
84
}

‎solution/2200-2299/2279.Maximum Bags With Full Capacity of Rocks/README.md‎

Lines changed: 58 additions & 60 deletions
Original file line numberDiff line numberDiff line change
@@ -74,6 +74,10 @@ tags:
7474

7575
### 方法一:排序 + 贪心
7676

77+
我们首先将每个背包的剩余容量计算出来,然后对剩余容量进行排序,接着我们从小到大遍历剩余容量,将额外的石头放入背包中,直到额外的石头用完或者背包的剩余容量用完为止,返回此时的背包数量即可。
78+
79+
时间复杂度 $O(n \times \log n),ドル空间复杂度 $O(\log n)$。其中 $n$ 为背包的数量。
80+
7781
<!-- tabs:start -->
7882

7983
#### Python3
@@ -83,37 +87,33 @@ class Solution:
8387
def maximumBags(
8488
self, capacity: List[int], rocks: List[int], additionalRocks: int
8589
) -> int:
86-
d = [a - b for a, b in zip(capacity, rocks)]
87-
d.sort()
88-
ans =0
89-
for v in d:
90-
if v <= additionalRocks:
91-
ans +=1
92-
additionalRocks -= v
93-
return ans
90+
for i, x in enumerate(rocks):
91+
capacity[i] -= x
92+
capacity.sort()
93+
for i, x in enumerate(capacity):
94+
additionalRocks -= x
95+
if additionalRocks <0:
96+
return i
97+
return len(capacity)
9498
```
9599

96100
#### Java
97101

98102
```java
99103
class Solution {
100104
public int maximumBags(int[] capacity, int[] rocks, int additionalRocks) {
101-
int n = capacity.length;
102-
int[] d = new int[n];
105+
int n = rocks.length;
103106
for (int i = 0; i < n; ++i) {
104-
d[i] =capacity[i] - rocks[i];
107+
capacity[i] -= rocks[i];
105108
}
106-
Arrays.sort(d);
107-
int ans = 0;
108-
for (int v : d) {
109-
if (v <= additionalRocks) {
110-
++ans;
111-
additionalRocks -= v;
112-
} else {
113-
break;
109+
Arrays.sort(capacity);
110+
for (int i = 0; i < n; ++i) {
111+
additionalRocks -= capacity[i];
112+
if (additionalRocks < 0) {
113+
return i;
114114
}
115115
}
116-
return ans;
116+
return n;
117117
}
118118
}
119119
```
@@ -124,17 +124,18 @@ class Solution {
124124
class Solution {
125125
public:
126126
int maximumBags(vector<int>& capacity, vector<int>& rocks, int additionalRocks) {
127-
int n = capacity.size();
128-
vector<int> d(n);
129-
for (int i = 0; i < n; ++i) d[i] = capacity[i] - rocks[i];
130-
sort(d.begin(), d.end());
131-
int ans = 0;
132-
for (int& v : d) {
133-
if (v > additionalRocks) break;
134-
++ans;
135-
additionalRocks -= v;
127+
int n = rocks.size();
128+
for (int i = 0; i < n; ++i) {
129+
capacity[i] -= rocks[i];
136130
}
137-
return ans;
131+
ranges::sort(capacity);
132+
for (int i = 0; i < n; ++i) {
133+
additionalRocks -= capacity[i];
134+
if (additionalRocks < 0) {
135+
return i;
136+
}
137+
}
138+
return n;
138139
}
139140
};
140141
```
@@ -143,58 +144,55 @@ public:
143144
144145
```go
145146
func maximumBags(capacity []int, rocks []int, additionalRocks int) int {
146-
n := len(capacity)
147-
d := make([]int, n)
148-
for i, v := range capacity {
149-
d[i] = v - rocks[i]
147+
for i, x := range rocks {
148+
capacity[i] -= x
150149
}
151-
sort.Ints(d)
152-
ans := 0
153-
for _, v := range d {
154-
if v > additionalRocks {
155-
break
150+
sort.Ints(capacity)
151+
for i, x := range capacity {
152+
additionalRocks -= x
153+
if additionalRocks < 0 {
154+
return i
156155
}
157-
ans++
158-
additionalRocks -= v
159156
}
160-
return ans
157+
return len(capacity)
161158
}
162159
```
163160

164161
#### TypeScript
165162

166163
```ts
167164
function maximumBags(capacity: number[], rocks: number[], additionalRocks: number): number {
168-
const n = capacity.length;
169-
const diffs = capacity.map((c, i) => c - rocks[i]);
170-
diffs.sort((a, b) => a - b);
171-
let ans = 0;
172-
for (let i = 0; i < n && (diffs[i] === 0 || diffs[i] <= additionalRocks); i++) {
173-
ans++;
174-
additionalRocks -= diffs[i];
165+
const n = rocks.length;
166+
for (let i = 0; i < n; ++i) {
167+
capacity[i] -= rocks[i];
168+
}
169+
capacity.sort((a, b) => a - b);
170+
for (let i = 0; i < n; ++i) {
171+
additionalRocks -= capacity[i];
172+
if (additionalRocks < 0) {
173+
return i;
174+
}
175175
}
176-
return ans;
176+
return n;
177177
}
178178
```
179179

180180
#### Rust
181181

182182
```rust
183183
impl Solution {
184-
pub fn maximum_bags(capacity: Vec<i32>, rocks: Vec<i32>, mut additional_rocks: i32) -> i32 {
185-
let n = capacity.len();
186-
let mut diffs = vec![0; n];
187-
for i in 0..n {
188-
diffs[i] = capacity[i] - rocks[i];
184+
pub fn maximum_bags(mut capacity: Vec<i32>, rocks: Vec<i32>, mut additional_rocks: i32) -> i32 {
185+
for i in 0..rocks.len() {
186+
capacity[i] -= rocks[i];
189187
}
190-
diffs.sort();
191-
for i in 0..n {
192-
if diffs[i] > additional_rocks {
188+
capacity.sort();
189+
for i in 0..capacity.len() {
190+
additional_rocks -= capacity[i];
191+
if additional_rocks < 0 {
193192
return i as i32;
194193
}
195-
additional_rocks -= diffs[i];
196194
}
197-
n as i32
195+
capacity.len() as i32
198196
}
199197
}
200198
```

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