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Commit 67fdde9

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feat: add solutions to lc problem: No.0043
No.0043.Multiply Strings
1 parent 8f9321e commit 67fdde9

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6 files changed

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-252
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6 files changed

+203
-252
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‎solution/0000-0099/0043.Multiply Strings/README.md‎

Lines changed: 67 additions & 68 deletions
Original file line numberDiff line numberDiff line change
@@ -38,6 +38,23 @@
3838

3939
<!-- 这里可写通用的实现逻辑 -->
4040

41+
**方法一:数学乘法模拟**
42+
43+
假设 `num1` 和 `num2` 的长度分别为 $m$ 和 $n,ドル则它们的乘积的长度最多为 $m + n$。
44+
45+
证明如下:
46+
47+
- 如果 `num1` 和 `num2` 都取最小值,那么它们的乘积为 ${10}^{m - 1} \times {10}^{n - 1} = {10}^{m + n - 2},ドル长度为 $m + n - 1$。
48+
- 如果 `num1` 和 `num2` 都取最大值,那么它们的乘积为 $({10}^m - 1) \times ({10}^n - 1) = {10}^{m + n} - {10}^m - {10}^n + 1,ドル长度为 $m + n$。
49+
50+
因此,我们可以申请一个长度为 $m + n$ 的数组,用于存储乘积的每一位。
51+
52+
从低位到高位,依次计算乘积的每一位,最后将数组转换为字符串即可。
53+
54+
注意判断最高位是否为 0ドル,ドル如果是,则去掉。
55+
56+
时间复杂度 $O(m \times n),ドル空间复杂度 $O(m + n)$。其中 $m$ 和 $n$ 分别为 `num1` 和 `num2` 的长度。
57+
4158
<!-- tabs:start -->
4259

4360
### **Python3**
@@ -47,27 +64,20 @@
4764
```python
4865
class Solution:
4966
def multiply(self, num1: str, num2: str) -> str:
50-
def mul(b, i):
51-
j, t = m - 1, 0
52-
while j >= 0 or t:
53-
if j >= 0:
54-
a = int(num1[j])
55-
t += a * b
56-
res[i] += t % 10
57-
if res[i] >= 10:
58-
res[i] %= 10
59-
res[i + 1] += 1
60-
i, j = i + 1, j - 1
61-
t //= 10
62-
67+
if num1 == "0" or num2 == "0":
68+
return "0"
6369
m, n = len(num1), len(num2)
64-
res = [0] * (m + n)
65-
for i in range(n):
66-
b = int(num2[n - 1 - i])
67-
mul(b, i)
68-
while len(res) > 1 and res[-1] == 0:
69-
res.pop()
70-
return ''.join([str(v) for v in res[::-1]])
70+
arr = [0] * (m + n)
71+
for i in range(m - 1, -1, -1):
72+
a = int(num1[i])
73+
for j in range(n - 1, -1, -1):
74+
b = int(num2[j])
75+
arr[i + j + 1] += a * b
76+
for i in range(m + n - 1, 0, -1):
77+
arr[i - 1] += arr[i] // 10
78+
arr[i] %= 10
79+
i = 0 if arr[0] else 1
80+
return "".join(str(x) for x in arr[i:])
7181
```
7282

7383
### **Java**
@@ -77,35 +87,28 @@ class Solution:
7787
```java
7888
class Solution {
7989
public String multiply(String num1, String num2) {
90+
if ("0".equals(num1) || "0".equals(num2)) {
91+
return "0";
92+
}
8093
int m = num1.length(), n = num2.length();
81-
int[] res = new int[m + n];
82-
for (int i = 0; i < n; ++i) {
83-
int b = num2.charAt(n - 1 - i) - '0';
84-
mul(num1, b, i, res);
94+
int[] arr = new int[m + n];
95+
for (int i = m - 1; i >= 0; --i) {
96+
int a = num1.charAt(i) - '0';
97+
for (int j = n - 1; j >= 0; --j) {
98+
int b = num2.charAt(j) - '0';
99+
arr[i + j + 1] += a * b;
100+
}
85101
}
86-
StringBuilder ans =newStringBuilder();
87-
for (int v : res) {
88-
ans.append(v);
102+
for (int i = arr.length -1; i >0; --i) {
103+
arr[i -1] += arr[i] /10;
104+
arr[i] %=10;
89105
}
90-
while (ans.length() > 1 && ans.charAt(ans.length() - 1) == '0') {
91-
ans.deleteCharAt(ans.length() - 1);
92-
}
93-
return ans.reverse().toString();
94-
}
95-
96-
private void mul(String A, int b, int i, int[] res) {
97-
for (int j = A.length() - 1, t = 0; j >= 0 || t > 0; --j) {
98-
if (j >= 0) {
99-
int a = A.charAt(j) - '0';
100-
t += a * b;
101-
}
102-
res[i++] += t % 10;
103-
if (res[i - 1] >= 10) {
104-
res[i - 1] %= 10;
105-
++res[i];
106-
}
107-
t /= 10;
106+
int i = arr[0] == 0 ? 1 : 0;
107+
StringBuilder ans = new StringBuilder();
108+
for (; i < arr.length; ++i) {
109+
ans.append(arr[i]);
108110
}
111+
return ans.toString();
109112
}
110113
}
111114
```
@@ -116,32 +119,28 @@ class Solution {
116119
class Solution {
117120
public:
118121
string multiply(string num1, string num2) {
119-
int m = num1.size(), n = num2.size();
120-
vector<int> res(m + n);
121-
for (int i = 0; i < n; ++i) {
122-
int b = num2[n - 1 - i] - '0';
123-
mul(num1, b, i, res);
122+
if (num1 == "0" || num2 == "0") {
123+
return "0";
124124
}
125-
string ans = "";
126-
for (int v : res) ans.push_back(v + '0');
127-
while (ans.size() > 1 && ans.back() == '0') ans.pop_back();
128-
reverse(ans.begin(), ans.end());
129-
return ans;
130-
}
131-
132-
void mul(string A, int b, int i, vector<int>& res) {
133-
for (int j = A.size() - 1, t = 0; j >= 0 || t > 0; --j) {
134-
if (j >= 0) {
135-
int a = A[j] - '0';
136-
t += a * b;
137-
}
138-
res[i++] += t % 10;
139-
if (res[i - 1] >= 10) {
140-
res[i - 1] %= 10;
141-
++res[i];
125+
int m = num1.size(), n = num2.size();
126+
vector<int> arr(m + n);
127+
for (int i = m - 1; i >= 0; --i) {
128+
int a = num1[i] - '0';
129+
for (int j = n - 1; j >= 0; --j) {
130+
int b = num2[j] - '0';
131+
arr[i + j + 1] += a * b;
142132
}
143-
t /= 10;
144133
}
134+
for (int i = arr.size() - 1; i; --i) {
135+
arr[i - 1] += arr[i] / 10;
136+
arr[i] %= 10;
137+
}
138+
int i = arr[0] ? 0 : 1;
139+
string ans;
140+
for (; i < arr.size(); ++i) {
141+
ans += '0' + arr[i];
142+
}
143+
return ans;
145144
}
146145
};
147146
```

‎solution/0000-0099/0043.Multiply Strings/README_EN.md‎

Lines changed: 68 additions & 92 deletions
Original file line numberDiff line numberDiff line change
@@ -34,63 +34,49 @@
3434
```python
3535
class Solution:
3636
def multiply(self, num1: str, num2: str) -> str:
37-
def mul(b, i):
38-
j, t = m - 1, 0
39-
while j >= 0 or t:
40-
if j >= 0:
41-
a = int(num1[j])
42-
t += a * b
43-
res[i] += t % 10
44-
if res[i] >= 10:
45-
res[i] %= 10
46-
res[i + 1] += 1
47-
i, j = i + 1, j - 1
48-
t //= 10
49-
37+
if num1 == "0" or num2 == "0":
38+
return "0"
5039
m, n = len(num1), len(num2)
51-
res = [0] * (m + n)
52-
for i in range(n):
53-
b = int(num2[n - 1 - i])
54-
mul(b, i)
55-
while len(res) > 1 and res[-1] == 0:
56-
res.pop()
57-
return ''.join([str(v) for v in res[::-1]])
40+
arr = [0] * (m + n)
41+
for i in range(m - 1, -1, -1):
42+
a = int(num1[i])
43+
for j in range(n - 1, -1, -1):
44+
b = int(num2[j])
45+
arr[i + j + 1] += a * b
46+
for i in range(m + n - 1, 0, -1):
47+
arr[i - 1] += arr[i] // 10
48+
arr[i] %= 10
49+
i = 0 if arr[0] else 1
50+
return "".join(str(x) for x in arr[i:])
5851
```
5952

6053
### **Java**
6154

6255
```java
6356
class Solution {
6457
public String multiply(String num1, String num2) {
65-
int m = num1.length(), n = num2.length();
66-
int[] res = new int[m + n];
67-
for (int i = 0; i < n; ++i) {
68-
int b = num2.charAt(n - 1 - i) - '0';
69-
mul(num1, b, i, res);
58+
if ("0".equals(num1) || "0".equals(num2)) {
59+
return "0";
7060
}
71-
StringBuilder ans = new StringBuilder();
72-
for (int v : res) {
73-
ans.append(v);
61+
int m = num1.length(), n = num2.length();
62+
int[] arr = new int[m + n];
63+
for (int i = m - 1; i >= 0; --i) {
64+
int a = num1.charAt(i) - '0';
65+
for (int j = n - 1; j >= 0; --j) {
66+
int b = num2.charAt(j) - '0';
67+
arr[i + j + 1] += a * b;
68+
}
7469
}
75-
while (ans.length() > 1 && ans.charAt(ans.length() - 1) == '0') {
76-
ans.deleteCharAt(ans.length() - 1);
70+
for (int i = arr.length - 1; i > 0; --i) {
71+
arr[i - 1] += arr[i] / 10;
72+
arr[i] %= 10;
7773
}
78-
return ans.reverse().toString();
79-
}
80-
81-
private void mul(String A, int b, int i, int[] res) {
82-
for (int j = A.length() - 1, t = 0; j >= 0 || t > 0; --j) {
83-
if (j >= 0) {
84-
int a = A.charAt(j) - '0';
85-
t += a * b;
86-
}
87-
res[i++] += t % 10;
88-
if (res[i - 1] >= 10) {
89-
res[i - 1] %= 10;
90-
++res[i];
91-
}
92-
t /= 10;
74+
int i = arr[0] == 0 ? 1 : 0;
75+
StringBuilder ans = new StringBuilder();
76+
for (; i < arr.length; ++i) {
77+
ans.append(arr[i]);
9378
}
79+
return ans.toString();
9480
}
9581
}
9682
```
@@ -101,32 +87,28 @@ class Solution {
10187
class Solution {
10288
public:
10389
string multiply(string num1, string num2) {
104-
int m = num1.size(), n = num2.size();
105-
vector<int> res(m + n);
106-
for (int i = 0; i < n; ++i) {
107-
int b = num2[n - 1 - i] - '0';
108-
mul(num1, b, i, res);
90+
if (num1 == "0" || num2 == "0") {
91+
return "0";
10992
}
110-
string ans = "";
111-
for (int v : res) ans.push_back(v + '0');
112-
while (ans.size() > 1 && ans.back() == '0') ans.pop_back();
113-
reverse(ans.begin(), ans.end());
114-
return ans;
115-
}
116-
117-
void mul(string A, int b, int i, vector<int>& res) {
118-
for (int j = A.size() - 1, t = 0; j >= 0 || t > 0; --j) {
119-
if (j >= 0) {
120-
int a = A[j] - '0';
121-
t += a * b;
122-
}
123-
res[i++] += t % 10;
124-
if (res[i - 1] >= 10) {
125-
res[i - 1] %= 10;
126-
++res[i];
93+
int m = num1.size(), n = num2.size();
94+
vector<int> arr(m + n);
95+
for (int i = m - 1; i >= 0; --i) {
96+
int a = num1[i] - '0';
97+
for (int j = n - 1; j >= 0; --j) {
98+
int b = num2[j] - '0';
99+
arr[i + j + 1] += a * b;
127100
}
128-
t /= 10;
129101
}
102+
for (int i = arr.size() - 1; i; --i) {
103+
arr[i - 1] += arr[i] / 10;
104+
arr[i] %= 10;
105+
}
106+
int i = arr[0] ? 0 : 1;
107+
string ans;
108+
for (; i < arr.size(); ++i) {
109+
ans += '0' + arr[i];
110+
}
111+
return ans;
130112
}
131113
};
132114
```
@@ -135,35 +117,29 @@ public:
135117
136118
```go
137119
func multiply(num1 string, num2 string) string {
120+
if num1 == "0" || num2 == "0" {
121+
return "0"
122+
}
138123
m, n := len(num1), len(num2)
139-
res := make([]int, m+n)
140-
mul := func(b, i int) {
141-
for j, t := m-1, 0; j >= 0 || t > 0; i, j = i+1, j-1 {
142-
if j >= 0 {
143-
a := int(num1[j] - '0')
144-
t += a * b
145-
}
146-
res[i] += t % 10
147-
if res[i] >= 10 {
148-
res[i] %= 10
149-
res[i+1]++
150-
}
151-
t /= 10
124+
arr := make([]int, m+n)
125+
for i := m - 1; i >= 0; i-- {
126+
a := int(num1[i] - '0')
127+
for j := n - 1; j >= 0; j-- {
128+
b := int(num2[j] - '0')
129+
arr[i+j+1] += a * b
152130
}
153131
}
154-
for i := 0; i < n; i++ {
155-
b := num2[n-1-i] - '0'
156-
mul(int(b), i)
157-
}
158-
var ans []byte
159-
for _, v := range res {
160-
ans = append(ans, byte(v+'0'))
132+
for i := len(arr) - 1; i > 0; i-- {
133+
arr[i-1] += arr[i] / 10
134+
arr[i] %= 10
161135
}
162-
for len(ans) > 1 && ans[len(ans)-1] == '0' {
163-
ans = ans[:len(ans)-1]
136+
i := 0
137+
if arr[0] == 0 {
138+
i = 1
164139
}
165-
for i, j := 0, len(ans)-1; i < j; i, j = i+1, j-1 {
166-
ans[i], ans[j] = ans[j], ans[i]
140+
ans := []byte{}
141+
for ; i < len(arr); i++ {
142+
ans = append(ans, byte('0'+arr[i]))
167143
}
168144
return string(ans)
169145
}

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