Skip to content

Navigation Menu

Sign in
Appearance settings

Search code, repositories, users, issues, pull requests...

Provide feedback

We read every piece of feedback, and take your input very seriously.

Saved searches

Use saved searches to filter your results more quickly
Sign up
Appearance settings

Commit 7c610c0

Browse files
feat: update solutions to lc problems: No.2262,2264 (#1719)
* No.2262.Total Appeal of A String * No.2264.Largest 3-Same-Digit Number in String
1 parent 9399197 commit 7c610c0

File tree

10 files changed

+155
-106
lines changed

10 files changed

+155
-106
lines changed

‎solution/2200-2299/2262.Total Appeal of A String/README.md‎

Lines changed: 20 additions & 12 deletions
Original file line numberDiff line numberDiff line change
@@ -56,14 +56,20 @@
5656

5757
<!-- 这里可写通用的实现逻辑 -->
5858

59-
遍历字符串 s 每个字符 `s[i]`, 维护以 `s[i]` 结尾的子串的引力值之和 t,遍历过程中累加 t 得到结果。
59+
**方法一:枚举**
6060

61-
若当前遍历到字符 `s[i]`,对应的引力值 t 的计算逻辑为:
61+
我们可以枚举以每个字符 $s[i]$ 结尾的字符串,计算其引力值之和 $t,ドル最后将所有 $t$ 相加即可。
6262

63-
1. 如果 `s[i]` 在之前没出现过,那么以 `s[i-1]` 结尾的每个子串的引力值都会加上 1,引力值之和会增加 i,再加上 1(`s[i]` 单独组成的子串的引力值),得到 `s[i]` 的引力值 t。
64-
1. 如果 `s[i]` 在之前出现过,定义最近一次出现的下标为 j,那么向子串 `s[0..i-1], s[1..i-1], ..., s[j..i-1]` 的末尾添加 `s[i]`,引力值不变。而 `s[j+1..i-1], s[j+2..i=1], ..., s[i-1..i-1]` 由于不包含 s[i],这些子串的引力值增加 1,因此有 i-j-1 个子串的引力值会增加 1,引力值之和增加 i-j-1,再加上 1,得到 `s[i]` 的引力值 t。
63+
考虑遍历到 $s[i]$ 时,即把 $s[i]$ 添加到以 $s[i-1]$ 结尾的子字符串的后面,其引力值之和 $t$ 的变化情况:
6564

66-
此过程中,我们用 pos 记录每个字符最近一次出现的位置。
65+
1. 如果 $s[i]$ 在之前没出现过,那么所有以 $s[i-1]$ 结尾的子字符串的引力值都会增加 1ドル,ドル共有 $i$ 个,所以 $t$ 增加 $i,ドル再加上 $s[i]$ 自身的引力值 1ドル,ドル所以 $t$ 一共增加 $i+1$;
66+
1. 如果 $s[i]$ 在之前出现过,不妨记上次出现的的位置为 $j,ドル那么我们向子字符串 $s[0..i-1],ドル $[1..i-1],ドル $s[2..i-1],ドル $\cdots,ドル $s[j..i-1]$ 后面添加 $s[i],ドル这些子字符串的引力值不会发生变化,因为 $s[i]$ 已经在这些子字符串中出现过了;而子字符串 $s[j+1..i-1],ドル $s[j+2..i-1],ドル $\cdots,ドル $s[i-1]$ 的引力值都会增加 1ドル,ドル共有 $i-j-1$ 个,所以 $t$ 增加 $i-j-1,ドル再加上 $s[i]$ 自身的引力值 1ドル,ドル所以 $t$ 一共增加 $i-j$。
67+
68+
综上,我们可以用一个数组 $pos$ 记录每个字符上次出现的位置,初始时所有位置都为 $-1,ドル
69+
70+
接下来,我们遍历字符串,每一次我们更新以当前字符结尾的子字符串的引力值之和 $t = t + i - pos[c],ドル其中 $c$ 是当前字符,累加 $t$ 到答案中。然后我们更新 $pos[c]$ 为当前位置 $i$。继续遍历直到字符串结束。
71+
72+
时间复杂度 $O(n),ドル空间复杂度 $O(|\Sigma|),ドル其中 $n$ 是字符串 $s$ 的长度;而 $|\Sigma|$ 是字符集的大小,本题中 $|\Sigma| = 26$。
6773

6874
<!-- tabs:start -->
6975

@@ -148,15 +154,17 @@ func appealSum(s string) int64 {
148154

149155
```ts
150156
function appealSum(s: string): number {
157+
const pos: number[] = Array(26).fill(-1);
151158
const n = s.length;
152-
let dp = new Array(n + 1).fill(0);
153-
const hashMap = new Map();
154-
for (let i = 0; i < n; i++) {
155-
const c = s.charAt(i);
156-
dp[i + 1] = dp[i] + i + 1 - (hashMap.get(c) || 0);
157-
hashMap.set(c, i + 1);
159+
let ans = 0;
160+
let t = 0;
161+
for (let i = 0; i < n; ++i) {
162+
const c = s.charCodeAt(i) - 97;
163+
t += i - pos[c];
164+
ans += t;
165+
pos[c] = i;
158166
}
159-
return dp.reduce((a, c) =>a+c, 0);
167+
return ans;
160168
}
161169
```
162170

‎solution/2200-2299/2262.Total Appeal of A String/README_EN.md‎

Lines changed: 24 additions & 7 deletions
Original file line numberDiff line numberDiff line change
@@ -52,6 +52,21 @@ The total sum is 4 +たす 6 +たす 6 +たす 4 =わ 20.
5252

5353
## Solutions
5454

55+
**Solution 1: Enumeration**
56+
57+
We can enumerate all the substrings that end with each character $s[i]$ and calculate their gravitational value sum $t$. Finally, we add up all the $t$ to get the total gravitational value sum.
58+
59+
When we reach $s[i],ドル which is added to the end of the substring that ends with $s[i-1],ドル we consider the change of the gravitational value sum $t$:
60+
61+
If $s[i]$ has not appeared before, then the gravitational value of all substrings that end with $s[i-1]$ will increase by 1ドル,ドル and there are a total of $i$ such substrings. Therefore, $t$ increases by $i,ドル plus the gravitational value of $s[i]$ itself, which is 1ドル$. Therefore, $t$ increases by a total of $i+1$.
62+
63+
If $s[i]$ has appeared before, let the last appearance position be $j$. Then we add $s[i]$ to the end of the substrings $s[0..i-1],ドル $[1..i-1],ドル $s[2..i-1],ドル $\cdots,ドル $s[j..i-1]$. The gravitational value of these substrings will not change because $s[i]$ has already appeared in these substrings. The gravitational value of the substrings $s[j+1..i-1],ドル $s[j+2..i-1],ドル $\cdots,ドル $s[i-1]$ will increase by 1ドル,ドル and there are a total of $i-j-1$ such substrings. Therefore, $t$ increases by $i-j-1,ドル plus the gravitational value of $s[i]$ itself, which is 1ドル$. Therefore, $t$ increases by a total of $i-j$.
64+
Therefore, we can use an array $pos$ to record the last appearance position of each character. Initially, all positions are set to $-1$.
65+
66+
Next, we traverse the string, and each time we update the gravitational value sum $t$ of the substring that ends with the current character to $t = t + i - pos[c],ドル where $c$ is the current character. We add $t$ to the answer. Then we update $pos[c]$ to the current position $i$. We continue to traverse until the end of the string.
67+
68+
The time complexity is $O(n),ドル and the space complexity is $O(|\Sigma|),ドル where $n$ is the length of the string $s,ドル and $|\Sigma|$ is the size of the character set. In this problem, $|\Sigma| = 26$.
69+
5570
<!-- tabs:start -->
5671

5772
### **Python3**
@@ -131,15 +146,17 @@ func appealSum(s string) int64 {
131146

132147
```ts
133148
function appealSum(s: string): number {
149+
const pos: number[] = Array(26).fill(-1);
134150
const n = s.length;
135-
let dp = new Array(n + 1).fill(0);
136-
const hashMap = new Map();
137-
for (let i = 0; i < n; i++) {
138-
const c = s.charAt(i);
139-
dp[i + 1] = dp[i] + i + 1 - (hashMap.get(c) || 0);
140-
hashMap.set(c, i + 1);
151+
let ans = 0;
152+
let t = 0;
153+
for (let i = 0; i < n; ++i) {
154+
const c = s.charCodeAt(i) - 97;
155+
t += i - pos[c];
156+
ans += t;
157+
pos[c] = i;
141158
}
142-
return dp.reduce((a, c) =>a+c, 0);
159+
return ans;
143160
}
144161
```
145162

Lines changed: 9 additions & 7 deletions
Original file line numberDiff line numberDiff line change
@@ -1,11 +1,13 @@
11
function appealSum(s: string): number {
2+
const pos: number[] = Array(26).fill(-1);
23
const n = s.length;
3-
let dp = new Array(n + 1).fill(0);
4-
const hashMap = new Map();
5-
for (let i = 0; i < n; i++) {
6-
const c = s.charAt(i);
7-
dp[i + 1] = dp[i] + i + 1 - (hashMap.get(c) || 0);
8-
hashMap.set(c, i + 1);
4+
let ans = 0;
5+
let t = 0;
6+
for (let i = 0; i < n; ++i) {
7+
const c = s.charCodeAt(i) - 97;
8+
t += i - pos[c];
9+
ans += t;
10+
pos[c] = i;
911
}
10-
return dp.reduce((a,c)=>a+c,0);
12+
return ans;
1113
}

‎solution/2200-2299/2264.Largest 3-Same-Digit Number in String/README.md‎

Lines changed: 34 additions & 24 deletions
Original file line numberDiff line numberDiff line change
@@ -62,6 +62,14 @@
6262

6363
<!-- 这里可写通用的实现逻辑 -->
6464

65+
**方法一:枚举**
66+
67+
我们可以从大到小枚举每个数字 $i,ドル其中 0ドル \le i \le 9,ドル然后判断连续的三个 $i$ 构成的字符串 $s$ 是否是 $num$ 的子串,若是,直接返回 $s$ 即可。
68+
69+
若枚举完所有的 $i$ 都没有找到满足条件的字符串,则返回空字符串。
70+
71+
时间复杂度 $O(10 \times n),ドル其中 $n$ 是字符串 $num$ 的长度。空间复杂度 $O(1)$。
72+
6573
<!-- tabs:start -->
6674

6775
### **Python3**
@@ -72,10 +80,9 @@
7280
class Solution:
7381
def largestGoodInteger(self, num: str) -> str:
7482
for i in range(9, -1, -1):
75-
t = str(i) * 3
76-
if t in num:
77-
return t
78-
return ''
83+
if (s := str(i) * 3) in num:
84+
return s
85+
return ""
7986
```
8087

8188
### **Java**
@@ -86,37 +93,27 @@ class Solution:
8693
class Solution {
8794
public String largestGoodInteger(String num) {
8895
for (int i = 9; i >= 0; i--) {
89-
String ret = String.valueOf(i).repeat(3);
90-
if (num.contains(ret)) {
91-
return ret;
96+
String s = String.valueOf(i).repeat(3);
97+
if (num.contains(s)) {
98+
return s;
9299
}
93100
}
94101
return "";
95102
}
96103
}
97104
```
98105

99-
### **TypeScript**
100-
101-
```ts
102-
function largestGoodInteger(num: string): string {
103-
for (let i = 9; i >= 0; i--) {
104-
const c = String(i).repeat(3);
105-
if (num.includes(c)) return c;
106-
}
107-
return '';
108-
}
109-
```
110-
111106
### **C++**
112107

113108
```cpp
114109
class Solution {
115110
public:
116111
string largestGoodInteger(string num) {
117112
for (char i = '9'; i >= '0'; --i) {
118-
string t(3, i);
119-
if (num.find(t) != string::npos) return t;
113+
string s(3, i);
114+
if (num.find(s) != string::npos) {
115+
return s;
116+
}
120117
}
121118
return "";
122119
}
@@ -128,15 +125,28 @@ public:
128125
```go
129126
func largestGoodInteger(num string) string {
130127
for c := '9'; c >= '0'; c-- {
131-
t := strings.Repeat(string(c), 3)
132-
if strings.Contains(num, t) {
133-
return t
128+
if s := strings.Repeat(string(c), 3); strings.Contains(num, s) {
129+
return s
134130
}
135131
}
136132
return ""
137133
}
138134
```
139135

136+
### **TypeScript**
137+
138+
```ts
139+
function largestGoodInteger(num: string): string {
140+
for (let i = 9; i >= 0; i--) {
141+
const s = String(i).repeat(3);
142+
if (num.includes(s)) {
143+
return s;
144+
}
145+
}
146+
return '';
147+
}
148+
```
149+
140150
### **...**
141151

142152
```

‎solution/2200-2299/2264.Largest 3-Same-Digit Number in String/README_EN.md‎

Lines changed: 34 additions & 24 deletions
Original file line numberDiff line numberDiff line change
@@ -56,6 +56,14 @@
5656

5757
## Solutions
5858

59+
**Solution 1: Enumeration**
60+
61+
We can enumerate each digit $i$ from large to small, where 0ドル \le i \le 9,ドル and then check whether the string $s$ consisting of three consecutive $i$ is a substring of $num$. If it is, we directly return $s$.
62+
63+
If we have enumerated all the possible values of $i$ and still haven't found a substring that satisfies the condition, we return an empty string.
64+
65+
The time complexity is $O(10 \times n),ドル where $n$ is the length of the string $num$. The space complexity is $O(1)$.
66+
5967
<!-- tabs:start -->
6068

6169
### **Python3**
@@ -64,10 +72,9 @@
6472
class Solution:
6573
def largestGoodInteger(self, num: str) -> str:
6674
for i in range(9, -1, -1):
67-
t = str(i) * 3
68-
if t in num:
69-
return t
70-
return ''
75+
if (s := str(i) * 3) in num:
76+
return s
77+
return ""
7178
```
7279

7380
### **Java**
@@ -76,37 +83,27 @@ class Solution:
7683
class Solution {
7784
public String largestGoodInteger(String num) {
7885
for (int i = 9; i >= 0; i--) {
79-
String ret = String.valueOf(i).repeat(3);
80-
if (num.contains(ret)) {
81-
return ret;
86+
String s = String.valueOf(i).repeat(3);
87+
if (num.contains(s)) {
88+
return s;
8289
}
8390
}
8491
return "";
8592
}
8693
}
8794
```
8895

89-
### **TypeScript**
90-
91-
```ts
92-
function largestGoodInteger(num: string): string {
93-
for (let i = 9; i >= 0; i--) {
94-
const c = String(i).repeat(3);
95-
if (num.includes(c)) return c;
96-
}
97-
return '';
98-
}
99-
```
100-
10196
### **C++**
10297

10398
```cpp
10499
class Solution {
105100
public:
106101
string largestGoodInteger(string num) {
107102
for (char i = '9'; i >= '0'; --i) {
108-
string t(3, i);
109-
if (num.find(t) != string::npos) return t;
103+
string s(3, i);
104+
if (num.find(s) != string::npos) {
105+
return s;
106+
}
110107
}
111108
return "";
112109
}
@@ -118,15 +115,28 @@ public:
118115
```go
119116
func largestGoodInteger(num string) string {
120117
for c := '9'; c >= '0'; c-- {
121-
t := strings.Repeat(string(c), 3)
122-
if strings.Contains(num, t) {
123-
return t
118+
if s := strings.Repeat(string(c), 3); strings.Contains(num, s) {
119+
return s
124120
}
125121
}
126122
return ""
127123
}
128124
```
129125

126+
### **TypeScript**
127+
128+
```ts
129+
function largestGoodInteger(num: string): string {
130+
for (let i = 9; i >= 0; i--) {
131+
const s = String(i).repeat(3);
132+
if (num.includes(s)) {
133+
return s;
134+
}
135+
}
136+
return '';
137+
}
138+
```
139+
130140
### **...**
131141

132142
```
Lines changed: 11 additions & 9 deletions
Original file line numberDiff line numberDiff line change
@@ -1,10 +1,12 @@
1-
class Solution {
2-
public:
3-
string largestGoodInteger(string num) {
4-
for (char i = '9'; i >= '0'; --i) {
5-
string t(3, i);
6-
if (num.find(t) != string::npos) return t;
7-
}
8-
return "";
9-
}
1+
class Solution {
2+
public:
3+
string largestGoodInteger(string num) {
4+
for (char i = '9'; i >= '0'; --i) {
5+
string s(3, i);
6+
if (num.find(s) != string::npos) {
7+
return s;
8+
}
9+
}
10+
return "";
11+
}
1012
};

‎solution/2200-2299/2264.Largest 3-Same-Digit Number in String/Solution.go‎

Lines changed: 2 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -1,8 +1,7 @@
11
func largestGoodInteger(num string) string {
22
for c := '9'; c >= '0'; c-- {
3-
t := strings.Repeat(string(c), 3)
4-
if strings.Contains(num, t) {
5-
return t
3+
if s := strings.Repeat(string(c), 3); strings.Contains(num, s) {
4+
return s
65
}
76
}
87
return ""
Lines changed: 11 additions & 11 deletions
Original file line numberDiff line numberDiff line change
@@ -1,11 +1,11 @@
1-
class Solution {
2-
public String largestGoodInteger(String num) {
3-
for (int i = 9; i >= 0; i--) {
4-
String ret = String.valueOf(i).repeat(3);
5-
if (num.contains(ret)) {
6-
return ret;
7-
}
8-
}
9-
return "";
10-
}
11-
}
1+
class Solution {
2+
public String largestGoodInteger(String num) {
3+
for (int i = 9; i >= 0; i--) {
4+
String s = String.valueOf(i).repeat(3);
5+
if (num.contains(s)) {
6+
return s;
7+
}
8+
}
9+
return "";
10+
}
11+
}

0 commit comments

Comments
(0)

AltStyle によって変換されたページ (->オリジナル) /