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Commit 07a0022

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feat: add solutions to lc problems: No.1903~1905 (#2102)
* No.1903.Largest Odd Number in String * No.1904.The Number of Full Rounds You Have Played * No.1905.Count Sub Islands
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‎solution/1900-1999/1903.Largest Odd Number in String/README.md‎

Lines changed: 35 additions & 19 deletions
Original file line numberDiff line numberDiff line change
@@ -49,7 +49,11 @@
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<!-- 这里可写通用的实现逻辑 -->
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52-
从后往前遍历字符串中的每个数字,遇到奇数则直接返回结果。若遍历结束仍未遇到奇数,返回空字符串。
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**方法一:逆序遍历**
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我们可以从后往前遍历字符串,找到第一个奇数,然后返回从开头到该奇数的子字符串即可。如果不存在奇数,则返回空字符串。
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时间复杂度 $O(n),ドル其中 $n$ 是字符串 $num$ 的长度。忽略答案字符串的空间消耗,空间复杂度 $O(1)$。
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<!-- tabs:start -->
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@@ -84,24 +88,6 @@ class Solution {
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}
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```
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### **JavaScript**
88-
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```js
90-
/**
91-
* @param {string} num
92-
* @return {string}
93-
*/
94-
var largestOddNumber = function (num) {
95-
let n = num.length;
96-
for (let j = n - 1; j >= 0; j--) {
97-
if (num.charAt(j) & (1 == 1)) {
98-
return num.slice(0, j + 1);
99-
}
100-
}
101-
return '';
102-
};
103-
```
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### **C++**
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```cpp
@@ -133,6 +119,36 @@ func largestOddNumber(num string) string {
133119
}
134120
```
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### **TypeScript**
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```ts
125+
function largestOddNumber(num: string): string {
126+
for (let i = num.length - 1; ~i; --i) {
127+
if (Number(num[i]) & 1) {
128+
return num.slice(0, i + 1);
129+
}
130+
}
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return '';
132+
}
133+
```
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### **JavaScript**
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```js
138+
/**
139+
* @param {string} num
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* @return {string}
141+
*/
142+
var largestOddNumber = function (num) {
143+
for (let i = num.length - 1; ~i; --i) {
144+
if (Number(num[i]) & 1) {
145+
return num.slice(0, i + 1);
146+
}
147+
}
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return '';
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};
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```
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### **...**
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```

‎solution/1900-1999/1903.Largest Odd Number in String/README_EN.md‎

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@@ -43,6 +43,12 @@
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## Solutions
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**Solution 1: Reverse Traversal**
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We can traverse the string from the end to the beginning, find the first odd number, and then return the substring from the beginning to this odd number. If there is no odd number, return an empty string.
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The time complexity is $O(n),ドル where $n$ is the length of the string $num$. Ignoring the space consumption of the answer string, the space complexity is $O(1)$.
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<!-- tabs:start -->
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### **Python3**
@@ -72,24 +78,6 @@ class Solution {
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}
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```
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### **JavaScript**
76-
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```js
78-
/**
79-
* @param {string} num
80-
* @return {string}
81-
*/
82-
var largestOddNumber = function (num) {
83-
let n = num.length;
84-
for (let j = n - 1; j >= 0; j--) {
85-
if (num.charAt(j) & (1 == 1)) {
86-
return num.slice(0, j + 1);
87-
}
88-
}
89-
return '';
90-
};
91-
```
92-
9381
### **C++**
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```cpp
@@ -121,6 +109,36 @@ func largestOddNumber(num string) string {
121109
}
122110
```
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112+
### **TypeScript**
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```ts
115+
function largestOddNumber(num: string): string {
116+
for (let i = num.length - 1; ~i; --i) {
117+
if (Number(num[i]) & 1) {
118+
return num.slice(0, i + 1);
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}
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}
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return '';
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}
123+
```
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### **JavaScript**
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```js
128+
/**
129+
* @param {string} num
130+
* @return {string}
131+
*/
132+
var largestOddNumber = function (num) {
133+
for (let i = num.length - 1; ~i; --i) {
134+
if (Number(num[i]) & 1) {
135+
return num.slice(0, i + 1);
136+
}
137+
}
138+
return '';
139+
};
140+
```
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### **...**
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```

‎solution/1900-1999/1903.Largest Odd Number in String/Solution.js‎

Lines changed: 3 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -3,10 +3,9 @@
33
* @return {string}
44
*/
55
var largestOddNumber = function (num) {
6-
let n = num.length;
7-
for (let j = n - 1; j >= 0; j--) {
8-
if (num.charAt(j) & (1 == 1)) {
9-
return num.slice(0, j + 1);
6+
for (let i = num.length - 1; ~i; --i) {
7+
if (Number(num[i]) & 1) {
8+
return num.slice(0, i + 1);
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}
1110
}
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return '';
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,8 @@
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function largestOddNumber(num: string): string {
2+
for (let i = num.length - 1; ~i; --i) {
3+
if (Number(num[i]) & 1) {
4+
return num.slice(0, i + 1);
5+
}
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}
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return '';
8+
}

‎solution/1900-1999/1904.The Number of Full Rounds You Have Played/README.md‎

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Original file line numberDiff line numberDiff line change
@@ -62,7 +62,13 @@
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<!-- 这里可写通用的实现逻辑 -->
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65-
转换为"分钟"进行计算。
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**方法一:转换为分钟数**
66+
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我们可以将输入的字符串转换为分钟数 $a$ 和 $b,ドル如果 $a > b,ドル则说明跨越了午夜,需要将 $b$ 加上一天的分钟数 1440ドル$。
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然后我们将 $a$ 向上取整到 15ドル$ 的倍数,将 $b$ 向下取整到 15ドル$ 的倍数,最后返回 $b$ 与 $a$ 的差值即可,注意要取 0ドル$ 和 $b - a$ 中的较大值。
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时间复杂度 $O(1),ドル空间复杂度 $O(1)$。
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<!-- tabs:start -->
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```python
7480
class Solution:
75-
def numberOfRounds(self, startTime: str, finishTime: str) -> int:
76-
def get(s: str) -> int:
81+
def numberOfRounds(self, loginTime: str, logoutTime: str) -> int:
82+
def f(s: str) -> int:
7783
return int(s[:2]) * 60 + int(s[3:])
7884

79-
start, finish = get(startTime), get(finishTime)
80-
if start > finish:
81-
finish += 24*60
82-
start, finish = (start + 14) // 15, finish // 15
83-
return max(0, finish - start)
85+
a, b = f(loginTime), f(logoutTime)
86+
if a > b:
87+
b += 1440
88+
a, b = (a + 14) // 15, b // 15
89+
return max(0, b - a)
8490
```
8591

8692
### **Java**
@@ -89,38 +95,19 @@ class Solution:
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9096
```java
9197
class Solution {
92-
public int numberOfRounds(String startTime, String finishTime) {
93-
int start = get(startTime), finish = get(finishTime);
94-
if (start > finish) {
95-
finish += 24*60;
98+
public int numberOfRounds(String loginTime, String logoutTime) {
99+
int a = f(loginTime), b = f(logoutTime);
100+
if (a > b) {
101+
b += 1440;
96102
}
97-
start = (start + 14) / 15;
98-
finish /= 15;
99-
return Math.max(0, finish - start);
103+
return Math.max(0, b / 15 - (a + 14) / 15);
100104
}
101105

102-
private int get(String s) {
103-
return Integer.parseInt(s.substring(0, 2)) * 60 + Integer.parseInt(s.substring(3));
104-
}
105-
}
106-
```
107-
108-
### **TypeScript**
109-
110-
```ts
111-
function numberOfRounds(startTime: string, finishTime: string): number {
112-
let m1 = toMinutes(startTime),
113-
m2 = toMinutes(finishTime);
114-
if (m1 > m2) {
115-
m2 += 24 * 60;
106+
private int f(String s) {
107+
int h = Integer.parseInt(s.substring(0, 2));
108+
int m = Integer.parseInt(s.substring(3, 5));
109+
return h * 60 + m;
116110
}
117-
let ans = Math.floor(m2 / 15) - Math.ceil(m1 / 15);
118-
return ans > 0 ? ans : 0;
119-
}
120-
121-
function toMinutes(time: string): number {
122-
let [h, m] = time.split(':').map(Number);
123-
return h * 60 + m;
124111
}
125112
```
126113

@@ -129,46 +116,51 @@ function toMinutes(time: string): number {
129116
```cpp
130117
class Solution {
131118
public:
132-
int numberOfRounds(string startTime, string finishTime) {
133-
int start = get(startTime), finish = get(finishTime);
134-
if (start > finish) {
135-
finish += 24 * 60;
119+
int numberOfRounds(string loginTime, string logoutTime) {
120+
auto f = [](string& s) {
121+
int h, m;
122+
sscanf(s.c_str(), "%d:%d", &h, &m);
123+
return h * 60 + m;
124+
};
125+
int a = f(loginTime), b = f(logoutTime);
126+
if (a > b) {
127+
b += 1440;
136128
}
137-
start = (start + 14) / 15;
138-
finish /= 15;
139-
return max(0, finish - start);
140-
}
141-
142-
private:
143-
int get(string s) {
144-
int a, b;
145-
sscanf(s.c_str(), "%d:%d", &a, &b);
146-
return a * 60 + b;
129+
return max(0, b / 15 - (a + 14) / 15);
147130
}
148131
};
149132
```
150133
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### **Go**
152135
153136
```go
154-
func numberOfRounds(startTime string, finishTime string) int {
155-
start, finish := get(startTime), get(finishTime)
156-
if start > finish {
157-
finish += 24 * 60
137+
func numberOfRounds(loginTime string, logoutTime string) int {
138+
f := func(s string) int {
139+
var h, m int
140+
fmt.Sscanf(s, "%d:%d", &h, &m)
141+
return h*60 + m
158142
}
159-
start = (start + 14) / 15
160-
finish /= 15
161-
if start > finish {
162-
return 0
143+
a, b := f(loginTime), f(logoutTime)
144+
if a > b {
145+
b += 1440
163146
}
164-
return finish - start
165-
147+
return max(0, b/15-(a+14)/15)
166148
}
149+
```
167150

168-
func get(s string) int {
169-
var a, b int
170-
fmt.Sscanf(s, "%d:%d", &a, &b)
171-
return a*60 + b
151+
### **TypeScript**
152+
153+
```ts
154+
function numberOfRounds(startTime: string, finishTime: string): number {
155+
const f = (s: string): number => {
156+
const [h, m] = s.split(':').map(Number);
157+
return h * 60 + m;
158+
};
159+
let [a, b] = [f(startTime), f(finishTime)];
160+
if (a > b) {
161+
b += 1440;
162+
}
163+
return Math.max(0, Math.floor(b / 15) - Math.ceil(a / 15));
172164
}
173165
```
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