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feat: add solutions to lc problems: No.0531,0533 (doocs#2628)
* No.0531.Lonely Pixel I * No.0533.Lonely Pixel II
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‎solution/0500-0599/0531.Lonely Pixel I/README.md

+63-46
Original file line numberDiff line numberDiff line change
@@ -42,28 +42,30 @@
4242

4343
## 解法
4444

45-
### 方法一
45+
### 方法一：计数 + 枚举
46+
47+
根据题目描述，我们需要统计每一行和每一列的黑色像素数量，分别记录在数组 $\textit{rows}$ 和 $\textit{cols}$ 中。然后我们遍历每一个黑色像素，检查其所在的行和列是否只有一个黑色像素，如果是则将答案加一。
48+
49+
时间复杂度 $O(m \times n)$，空间复杂度 $O(m + n)$。其中 $m$ 和 $n$ 分别是矩阵的行数和列数。
4650

4751
<!-- tabs:start -->
4852

4953
```python
5054
class Solution:
5155
def findLonelyPixel(self, picture: List[List[str]]) -> int:
52-
m, n = len(picture), len(picture[0])
53-
rows, cols = [0] * m, [0] * n
54-
for i in range(m):
55-
for j in range(n):
56-
if picture[i][j] == 'B':
56+
rows = [0] * len(picture)
57+
cols = [0] * len(picture[0])
58+
for i, row in enumerate(picture):
59+
for j, x in enumerate(row):
60+
if x == "B":
5761
rows[i] += 1
5862
cols[j] += 1
59-
res = 0
60-
for i in range(m):
61-
if rows[i] == 1:
62-
for j in range(n):
63-
if picture[i][j] == 'B' and cols[j] == 1:
64-
res += 1
65-
break
66-
return res
63+
ans = 0
64+
for i, row in enumerate(picture):
65+
for j, x in enumerate(row):
66+
if x == "B" and rows[i] == 1 and cols[j] == 1:
67+
ans += 1
68+
return ans
6769
```
6870

6971
```java
@@ -80,18 +82,15 @@ class Solution {
8082
}
8183
}
8284
}
83-
int res = 0;
85+
int ans = 0;
8486
for (int i = 0; i < m; ++i) {
85-
if (rows[i] == 1) {
86-
for (int j = 0; j < n; ++j) {
87-
if (picture[i][j] == 'B' && cols[j] == 1) {
88-
++res;
89-
break;
90-
}
87+
for (int j = 0; j < n; ++j) {
88+
if (picture[i][j] == 'B' && rows[i] == 1 && cols[j] == 1) {
89+
++ans;
9190
}
9291
}
9392
}
94-
return res;
93+
return ans;
9594
}
9695
}
9796
```
@@ -111,47 +110,65 @@ public:
111110
}
112111
}
113112
}
114-
int res = 0;
113+
int ans = 0;
115114
for (int i = 0; i < m; ++i) {
116-
if (rows[i] == 1) {
117-
for (int j = 0; j < n; ++j) {
118-
if (picture[i][j] == 'B' && cols[j] == 1) {
119-
++res;
120-
break;
121-
}
115+
for (int j = 0; j < n; ++j) {
116+
if (picture[i][j] == 'B' && rows[i] == 1 && cols[j] == 1) {
117+
++ans;
122118
}
123119
}
124120
}
125-
return res;
121+
return ans;
126122
}
127123
};
128124
```
129125
130126
```go
131-
func findLonelyPixel(picture [][]byte) int {
132-
m, n := len(picture), len(picture[0])
133-
rows := make([]int, m)
134-
cols := make([]int, n)
135-
for i := 0; i < m; i++ {
136-
for j := 0; j < n; j++ {
137-
if picture[i][j] == 'B' {
127+
func findLonelyPixel(picture [][]byte) (ans int) {
128+
rows := make([]int, len(picture))
129+
cols := make([]int, len(picture[0]))
130+
for i, row := range picture {
131+
for j, x := range row {
132+
if x == 'B' {
138133
rows[i]++
139134
cols[j]++
140135
}
141136
}
142137
}
143-
res := 0
144-
for i := 0; i < m; i++ {
145-
if rows[i] == 1 {
146-
for j := 0; j < n; j++ {
147-
if picture[i][j] == 'B' && cols[j] == 1 {
148-
res++
149-
break
150-
}
138+
for i, row := range picture {
139+
for j, x := range row {
140+
if x == 'B' && rows[i] == 1 && cols[j] == 1 {
141+
ans++
151142
}
152143
}
153144
}
154-
return res
145+
return
146+
}
147+
```
148+
149+
```ts
150+
function findLonelyPixel(picture: string[][]): number {
151+
const m = picture.length;
152+
const n = picture[0].length;
153+
const rows: number[] = Array(m).fill(0);
154+
const cols: number[] = Array(n).fill(0);
155+
for (let i = 0; i < m; ++i) {
156+
for (let j = 0; j < n; ++j) {
157+
if (picture[i][j] === 'B') {
158+
++rows[i];
159+
++cols[j];
160+
}
161+
}
162+
}
163+
let ans = 0;
164+
for (let i = 0; i < m; ++i) {
165+
for (let j = 0; j < n; ++j) {
166+
if (picture[i][j] === 'B' && rows[i] === 1 && cols[j] === 1) {
167+
++ans;
168+
}
169+
}
170+
}
171+
return ans;
155172
}
156173
```
157174

‎solution/0500-0599/0531.Lonely Pixel I/README_EN.md

+63-46
Original file line numberDiff line numberDiff line change
@@ -38,28 +38,30 @@
3838

3939
## Solutions
4040

41-
### Solution 1
41+
### Solution 1: Counting + Enumeration
42+
43+
According to the problem description, we need to count the number of black pixels in each row and column, which are recorded in the arrays `rows` and `cols` respectively. Then we traverse each black pixel, check whether there is only one black pixel in its row and column. If so, we increment the answer by one.
44+
45+
The time complexity is $O(m \times n)$, and the space complexity is $O(m + n)$, where $m$ and $n$ are the number of rows and columns in the matrix respectively.
4246

4347
<!-- tabs:start -->
4448

4549
```python
4650
class Solution:
4751
def findLonelyPixel(self, picture: List[List[str]]) -> int:
48-
m, n = len(picture), len(picture[0])
49-
rows, cols = [0] * m, [0] * n
50-
for i in range(m):
51-
for j in range(n):
52-
if picture[i][j] == 'B':
52+
rows = [0] * len(picture)
53+
cols = [0] * len(picture[0])
54+
for i, row in enumerate(picture):
55+
for j, x in enumerate(row):
56+
if x == "B":
5357
rows[i] += 1
5458
cols[j] += 1
55-
res = 0
56-
for i in range(m):
57-
if rows[i] == 1:
58-
for j in range(n):
59-
if picture[i][j] == 'B' and cols[j] == 1:
60-
res += 1
61-
break
62-
return res
59+
ans = 0
60+
for i, row in enumerate(picture):
61+
for j, x in enumerate(row):
62+
if x == "B" and rows[i] == 1 and cols[j] == 1:
63+
ans += 1
64+
return ans
6365
```
6466

6567
```java
@@ -76,18 +78,15 @@ class Solution {
7678
}
7779
}
7880
}
79-
int res = 0;
81+
int ans = 0;
8082
for (int i = 0; i < m; ++i) {
81-
if (rows[i] == 1) {
82-
for (int j = 0; j < n; ++j) {
83-
if (picture[i][j] == 'B' && cols[j] == 1) {
84-
++res;
85-
break;
86-
}
83+
for (int j = 0; j < n; ++j) {
84+
if (picture[i][j] == 'B' && rows[i] == 1 && cols[j] == 1) {
85+
++ans;
8786
}
8887
}
8988
}
90-
return res;
89+
return ans;
9190
}
9291
}
9392
```
@@ -107,47 +106,65 @@ public:
107106
}
108107
}
109108
}
110-
int res = 0;
109+
int ans = 0;
111110
for (int i = 0; i < m; ++i) {
112-
if (rows[i] == 1) {
113-
for (int j = 0; j < n; ++j) {
114-
if (picture[i][j] == 'B' && cols[j] == 1) {
115-
++res;
116-
break;
117-
}
111+
for (int j = 0; j < n; ++j) {
112+
if (picture[i][j] == 'B' && rows[i] == 1 && cols[j] == 1) {
113+
++ans;
118114
}
119115
}
120116
}
121-
return res;
117+
return ans;
122118
}
123119
};
124120
```
125121
126122
```go
127-
func findLonelyPixel(picture [][]byte) int {
128-
m, n := len(picture), len(picture[0])
129-
rows := make([]int, m)
130-
cols := make([]int, n)
131-
for i := 0; i < m; i++ {
132-
for j := 0; j < n; j++ {
133-
if picture[i][j] == 'B' {
123+
func findLonelyPixel(picture [][]byte) (ans int) {
124+
rows := make([]int, len(picture))
125+
cols := make([]int, len(picture[0]))
126+
for i, row := range picture {
127+
for j, x := range row {
128+
if x == 'B' {
134129
rows[i]++
135130
cols[j]++
136131
}
137132
}
138133
}
139-
res := 0
140-
for i := 0; i < m; i++ {
141-
if rows[i] == 1 {
142-
for j := 0; j < n; j++ {
143-
if picture[i][j] == 'B' && cols[j] == 1 {
144-
res++
145-
break
146-
}
134+
for i, row := range picture {
135+
for j, x := range row {
136+
if x == 'B' && rows[i] == 1 && cols[j] == 1 {
137+
ans++
147138
}
148139
}
149140
}
150-
return res
141+
return
142+
}
143+
```
144+
145+
```ts
146+
function findLonelyPixel(picture: string[][]): number {
147+
const m = picture.length;
148+
const n = picture[0].length;
149+
const rows: number[] = Array(m).fill(0);
150+
const cols: number[] = Array(n).fill(0);
151+
for (let i = 0; i < m; ++i) {
152+
for (let j = 0; j < n; ++j) {
153+
if (picture[i][j] === 'B') {
154+
++rows[i];
155+
++cols[j];
156+
}
157+
}
158+
}
159+
let ans = 0;
160+
for (let i = 0; i < m; ++i) {
161+
for (let j = 0; j < n; ++j) {
162+
if (picture[i][j] === 'B' && rows[i] === 1 && cols[j] === 1) {
163+
++ans;
164+
}
165+
}
166+
}
167+
return ans;
151168
}
152169
```
153170

‎solution/0500-0599/0531.Lonely Pixel I/Solution.cpp

+5-8
Original file line numberDiff line numberDiff line change
@@ -12,17 +12,14 @@ class Solution {
1212
}
1313
}
1414
}
15-
int res = 0;
15+
int ans = 0;
1616
for (int i = 0; i < m; ++i) {
17-
if (rows[i] == 1) {
18-
for (int j = 0; j < n; ++j) {
19-
if (picture[i][j] == 'B' && cols[j] == 1) {
20-
++res;
21-
break;
22-
}
17+
for (int j = 0; j < n; ++j) {
18+
if (picture[i][j] == 'B' && rows[i] == 1 && cols[j] == 1) {
19+
++ans;
2320
}
2421
}
2522
}
26-
return res;
23+
return ans;
2724
}
2825
};

‎solution/0500-0599/0531.Lonely Pixel I/Solution.go

+11-16
Original file line numberDiff line numberDiff line change
@@ -1,25 +1,20 @@
1-
func findLonelyPixel(picture [][]byte) int {
2-
m, n := len(picture), len(picture[0])
3-
rows := make([]int, m)
4-
cols := make([]int, n)
5-
for i := 0; i < m; i++ {
6-
for j := 0; j < n; j++ {
7-
if picture[i][j] == 'B' {
1+
func findLonelyPixel(picture [][]byte) (ans int) {
2+
rows := make([]int, len(picture))
3+
cols := make([]int, len(picture[0]))
4+
for i, row := range picture {
5+
for j, x := range row {
6+
if x == 'B' {
87
rows[i]++
98
cols[j]++
109
}
1110
}
1211
}
13-
res := 0
14-
for i := 0; i < m; i++ {
15-
if rows[i] == 1 {
16-
for j := 0; j < n; j++ {
17-
if picture[i][j] == 'B' && cols[j] == 1 {
18-
res++
19-
break
20-
}
12+
for i, row := range picture {
13+
for j, x := range row {
14+
if x == 'B' && rows[i] == 1 && cols[j] == 1 {
15+
ans++
2116
}
2217
}
2318
}
24-
return res
19+
return
2520
}

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