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Jul 6, 2024
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feat: update solutions to lc problem: No.0942 #3210

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68 changes: 35 additions & 33 deletions solution/0900-0999/0942.DI String Match/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -65,7 +65,13 @@ tags:

<!-- solution:start -->

### 方法一
### 方法一：贪心

我们可以使用两个指针 `low` 和 `high` 分别表示当前的最小值和最大值，然后遍历字符串 `s`，如果当前字符是 `I`，那么我们就将 `low` 加入到结果数组中，并且 `low` 自增 1；如果当前字符是 `D`，那么我们就将 `high` 加入到结果数组中，并且 `high` 自减 1。

最后，我们将 `low` 加入到结果数组中，返回结果数组即可。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为字符串 `s` 的长度。

<!-- tabs:start -->

Expand All @@ -74,11 +80,10 @@ tags:
```python
class Solution:
def diStringMatch(self, s: str) -> List[int]:
n = len(s)
low, high = 0, n
low, high = 0, len(s)
ans = []
for i in range(n):
if s[i] == 'I':
for c in s:
if c == "I":
ans.append(low)
low += 1
else:
Expand Down Expand Up @@ -134,12 +139,10 @@ public:
#### Go

```go
func diStringMatch(s string) []int {
n := len(s)
low, high := 0, n
var ans []int
for i := 0; i < n; i++ {
if s[i] == 'I' {
func diStringMatch(s string) (ans []int) {
low, high := 0, len(s)
for _, c := range s {
if c == 'I' {
ans = append(ans, low)
low++
} else {
Expand All @@ -148,27 +151,25 @@ func diStringMatch(s string) []int {
}
}
ans = append(ans, low)
return ans
return
}
```

#### TypeScript

```ts
function diStringMatch(s: string): number[] {
const n = s.length;
const res = new Array(n + 1);
let low = 0;
let high = n;
for (let i = 0; i < n; i++) {
if (s[i] === 'I') {
res[i] = low++;
const ans: number[] = [];
let [low, high] = [0, s.length];
for (const c of s) {
if (c === 'I') {
ans.push(low++);
} else {
res[i] = high--;
ans.push(high--);
}
}
res[n] = low;
return res;
ans.push(low);
return ans;
}
```

Expand All @@ -177,21 +178,22 @@ function diStringMatch(s: string): number[] {
```rust
impl Solution {
pub fn di_string_match(s: String) -> Vec<i32> {
let s = s.as_bytes();
let n = s.len();
let mut res = Vec::with_capacity(n + 1);
let (mut low, mut high) = (-1, (n + 1) as i32);
for i in 0..n {
res.push(if s[i] == b'I' {
let mut low = 0;
let mut high = s.len() as i32;
let mut ans = Vec::with_capacity(s.len() + 1);

for c in s.chars() {
if c == 'I' {
ans.push(low);
low += 1;
low
} else {
ans.push(high);
high -= 1;
high
});
}
}
res.push(low + 1);
res

ans.push(low);
ans
}
}
```
Expand Down
68 changes: 35 additions & 33 deletions solution/0900-0999/0942.DI String Match/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -53,7 +53,13 @@ tags:

<!-- solution:start -->

### Solution 1
### Solution 1: Greedy Algorithm

We can use two pointers `low` and `high` to represent the current minimum and maximum values, respectively. Then, we traverse the string `s`. If the current character is `I`, we add `low` to the result array, and increment `low` by 1; if the current character is `D`, we add `high` to the result array, and decrement `high` by 1.

Finally, we add `low` to the result array and return the result array.

The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the string `s`.

<!-- tabs:start -->

Expand All @@ -62,11 +68,10 @@ tags:
```python
class Solution:
def diStringMatch(self, s: str) -> List[int]:
n = len(s)
low, high = 0, n
low, high = 0, len(s)
ans = []
for i in range(n):
if s[i] == 'I':
for c in s:
if c == "I":
ans.append(low)
low += 1
else:
Expand Down Expand Up @@ -122,12 +127,10 @@ public:
#### Go

```go
func diStringMatch(s string) []int {
n := len(s)
low, high := 0, n
var ans []int
for i := 0; i < n; i++ {
if s[i] == 'I' {
func diStringMatch(s string) (ans []int) {
low, high := 0, len(s)
for _, c := range s {
if c == 'I' {
ans = append(ans, low)
low++
} else {
Expand All @@ -136,27 +139,25 @@ func diStringMatch(s string) []int {
}
}
ans = append(ans, low)
return ans
return
}
```

#### TypeScript

```ts
function diStringMatch(s: string): number[] {
const n = s.length;
const res = new Array(n + 1);
let low = 0;
let high = n;
for (let i = 0; i < n; i++) {
if (s[i] === 'I') {
res[i] = low++;
const ans: number[] = [];
let [low, high] = [0, s.length];
for (const c of s) {
if (c === 'I') {
ans.push(low++);
} else {
res[i] = high--;
ans.push(high--);
}
}
res[n] = low;
return res;
ans.push(low);
return ans;
}
```

Expand All @@ -165,21 +166,22 @@ function diStringMatch(s: string): number[] {
```rust
impl Solution {
pub fn di_string_match(s: String) -> Vec<i32> {
let s = s.as_bytes();
let n = s.len();
let mut res = Vec::with_capacity(n + 1);
let (mut low, mut high) = (-1, (n + 1) as i32);
for i in 0..n {
res.push(if s[i] == b'I' {
let mut low = 0;
let mut high = s.len() as i32;
let mut ans = Vec::with_capacity(s.len() + 1);

for c in s.chars() {
if c == 'I' {
ans.push(low);
low += 1;
low
} else {
ans.push(high);
high -= 1;
high
});
}
}
res.push(low + 1);
res

ans.push(low);
ans
}
}
```
Expand Down
12 changes: 5 additions & 7 deletions solution/0900-0999/0942.DI String Match/Solution.go
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Open in desktop
Original file line number Diff line number Diff line change
@@ -1,9 +1,7 @@
func diStringMatch(s string) []int {
n := len(s)
low, high := 0, n
var ans []int
for i := 0; i < n; i++ {
if s[i] == 'I' {
func diStringMatch(s string) (ans []int) {
low, high := 0, len(s)
for _, c := range s {
if c == 'I' {
ans = append(ans, low)
low++
} else {
Expand All @@ -12,5 +10,5 @@ func diStringMatch(s string) []int {
}
}
ans = append(ans, low)
return ans
return
}
7 changes: 3 additions & 4 deletions solution/0900-0999/0942.DI String Match/Solution.py
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,10 +1,9 @@
class Solution:
def diStringMatch(self, s: str) -> List[int]:
n = len(s)
low, high = 0, n
low, high = 0, len(s)
ans = []
for i in range(n):
if s[i] == 'I':
for c in s:
if c == "I":
ans.append(low)
low += 1
else:
Expand Down
23 changes: 12 additions & 11 deletions solution/0900-0999/0942.DI String Match/Solution.rs
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,19 +1,20 @@
impl Solution {
pub fn di_string_match(s: String) -> Vec<i32> {
let s = s.as_bytes();
let n = s.len();
let mut res = Vec::with_capacity(n + 1);
let (mut low, mut high) = (-1, (n + 1) as i32);
for i in 0..n {
res.push(if s[i] == b'I' {
let mut low = 0;
let mut high = s.len() as i32;
let mut ans = Vec::with_capacity(s.len() + 1);

for c in s.chars() {
if c == 'I' {
ans.push(low);
low += 1;
low
} else {
ans.push(high);
high -= 1;
high
});
}
}
res.push(low + 1);
res

ans.push(low);
ans
}
}
18 changes: 8 additions & 10 deletions solution/0900-0999/0942.DI String Match/Solution.ts
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,15 +1,13 @@
function diStringMatch(s: string): number[] {
const n = s.length;
const res = new Array(n + 1);
let low = 0;
let high = n;
for (let i = 0; i < n; i++) {
if (s[i] === 'I') {
res[i] = low++;
const ans: number[] = [];
let [low, high] = [0, s.length];
for (const c of s) {
if (c === 'I') {
ans.push(low++);
} else {
res[i] = high--;
ans.push(high--);
}
}
res[n] = low;
return res;
ans.push(low);
return ans;
}
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