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Sep 24, 2025
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feat: add solutions to lc problem: No.0120 #4747

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119 changes: 51 additions & 68 deletions solution/0100-0199/0120.Triangle/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -73,17 +73,11 @@ tags:
我们定义 $f[i][j]$ 表示从三角形底部走到位置 $(i, j)$ 的最小路径和。这里的位置 $(i, j)$ 指的是三角形中第 $i$ 行第 $j$ 列（均从 $0$ 开始编号）的位置。那么我们有如下的状态转移方程：

$$
f[i][j] = \min(f[i + 1][j], f[i + 1][j + 1]) + triangle[i][j]
f[i][j] = \min(f[i + 1][j], f[i + 1][j + 1]) + \text{triangle}[i][j]
$$

答案即为 $f[0][0]$。

我们注意到，状态 $f[i][j]$ 仅与状态 $f[i + 1][j]$ 和状态 $f[i + 1][j + 1]$ 有关，因此我们可以使用一维数组代替二维数组，将空间复杂度从 $O(n^2)$ 降低至 $O(n)$。

时间复杂度 $O(n^2)$，空间复杂度 $O(n)$。其中 $n$ 是三角形的行数。

更进一步，我们还可以直接复用 $triangle$ 作为 $f$ 数组，这样就无需再额外创建 $f$ 数组，空间复杂度降低至 $O(1)$。

<!-- tabs:start -->

#### Python3
Expand All @@ -105,13 +99,13 @@ class Solution:
class Solution {
public int minimumTotal(List<List<Integer>> triangle) {
int n = triangle.size();
int[] f = new int[n + 1];
int[][] f = new int[n + 1][n + 1];
for (int i = n - 1; i >= 0; --i) {
for (int j = 0; j <= i; ++j) {
f[j] = Math.min(f[j], f[j + 1]) + triangle.get(i).get(j);
f[i][j] = Math.min(f[i + 1][j], f[i + 1][j + 1]) + triangle.get(i).get(j);
}
}
return f[0];
return f[0][0];
}
}
```
Expand All @@ -123,14 +117,13 @@ class Solution {
public:
int minimumTotal(vector<vector<int>>& triangle) {
int n = triangle.size();
int f[n + 1];
memset(f, 0, sizeof(f));
for (int i = n - 1; ~i; --i) {
vector<vector<int>> f(n + 1, vector<int>(n + 1, 0));
for (int i = n - 1; i >= 0; --i) {
for (int j = 0; j <= i; ++j) {
f[j] = min(f[j], f[j + 1]) + triangle[i][j];
f[i][j] = min(f[i + 1][j], f[i + 1][j + 1]) + triangle[i][j];
}
}
return f[0];
return f[0][0];
}
};
```
Expand All @@ -140,13 +133,16 @@ public:
```go
func minimumTotal(triangle [][]int) int {
n := len(triangle)
f := make([]int, n+1)
f := make([][]int, n+1)
for i := range f {
f[i] = make([]int, n+1)
}
for i := n - 1; i >= 0; i-- {
for j := 0; j <= i; j++ {
f[j] = min(f[j], f[j+1]) + triangle[i][j]
f[i][j] = min(f[i+1][j], f[i+1][j+1]) + triangle[i][j]
}
}
return f[0]
return f[0][0]
}
```

Expand All @@ -155,13 +151,13 @@ func minimumTotal(triangle [][]int) int {
```ts
function minimumTotal(triangle: number[][]): number {
const n = triangle.length;
const f: number[] = Array(n + 1).fill(0);
for (let i = n - 1; ~i; --i) {
const f: number[][] = Array.from({ length: n + 1 }, () => Array(n + 1).fill(0));
for (let i = n - 1; i >= 0; --i) {
for (let j = 0; j <= i; ++j) {
f[j] = Math.min(f[j], f[j + 1]) + triangle[i][j];
f[i][j] = Math.min(f[i + 1][j], f[i + 1][j + 1]) + triangle[i][j];
}
}
return f[0];
return f[0][0];
}
```

Expand All @@ -171,13 +167,13 @@ function minimumTotal(triangle: number[][]): number {
impl Solution {
pub fn minimum_total(triangle: Vec<Vec<i32>>) -> i32 {
let n = triangle.len();
let mut f = vec![0; n + 1];
let mut f = vec![vec![0; n + 1]; n + 1];
for i in (0..n).rev() {
for j in 0..=i {
f[j] = f[j].min(f[j + 1]) + triangle[i][j];
f[i][j] = f[i + 1][j].min(f[i + 1][j + 1]) + triangle[i][j];
}
}
f[0]
f[0][0]
}
}
```
Expand All @@ -188,7 +184,11 @@ impl Solution {

<!-- solution:start -->

### 方法二
### 方法二：动态规划（空间优化）

我们注意到，状态 $f[i][j]$ 仅与状态 $f[i + 1][j]$ 和状态 $f[i + 1][j + 1]$ 有关，因此我们可以使用一维数组代替二维数组，将空间复杂度从 $O(n^2)$ 降低至 $O(n)$。

时间复杂度 $O(n^2)$，空间复杂度 $O(n)$。其中 $n$ 是三角形的行数。

<!-- tabs:start -->

Expand All @@ -210,14 +210,14 @@ class Solution:
```java
class Solution {
public int minimumTotal(List<List<Integer>> triangle) {
for (int i = triangle.size() - 2; i >= 0; --i) {
int n = triangle.size();
int[] f = new int[n + 1];
for (int i = n - 1; i >= 0; --i) {
for (int j = 0; j <= i; ++j) {
int x = triangle.get(i).get(j);
int y = Math.min(triangle.get(i + 1).get(j), triangle.get(i + 1).get(j + 1));
triangle.get(i).set(j, x + y);
f[j] = Math.min(f[j], f[j + 1]) + triangle.get(i).get(j);
}
}
return triangle.get(0).get(0);
return f[0];
}
}
```
Expand All @@ -228,12 +228,14 @@ class Solution {
class Solution {
public:
int minimumTotal(vector<vector<int>>& triangle) {
for (int i = triangle.size() - 2; ~i; --i) {
int n = triangle.size();
vector<int> f(n + 1, 0);
for (int i = n - 1; i >= 0; --i) {
for (int j = 0; j <= i; ++j) {
triangle[i][j] += min(triangle[i + 1][j], triangle[i + 1][j + 1]);
f[j] = min(f[j], f[j + 1]) + triangle[i][j];
}
}
return triangle[0][0];
return f[0];
}
};
```
Expand All @@ -242,25 +244,29 @@ public:

```go
func minimumTotal(triangle [][]int) int {
for i := len(triangle) - 2; i >= 0; i-- {
n := len(triangle)
f := make([]int, n+1)
for i := n - 1; i >= 0; i-- {
for j := 0; j <= i; j++ {
triangle[i][j] += min(triangle[i+1][j], triangle[i+1][j+1])
f[j] = min(f[j], f[j+1]) + triangle[i][j]
}
}
return triangle[0][0]
return f[0]
}
```

#### TypeScript

```ts
function minimumTotal(triangle: number[][]): number {
for (let i = triangle.length - 2; ~i; --i) {
const n = triangle.length;
const f: number[] = Array(n + 1).fill(0);
for (let i = n - 1; i >= 0; --i) {
for (let j = 0; j <= i; ++j) {
triangle[i][j] += Math.min(triangle[i + 1][j], triangle[i + 1][j + 1]);
f[j] = Math.min(f[j], f[j + 1]) + triangle[i][j];
}
}
return triangle[0][0];
return f[0];
}
```

Expand All @@ -269,13 +275,14 @@ function minimumTotal(triangle: number[][]): number {
```rust
impl Solution {
pub fn minimum_total(triangle: Vec<Vec<i32>>) -> i32 {
let mut triangle = triangle;
for i in (0..triangle.len() - 1).rev() {
let n = triangle.len();
let mut f = vec![0; n + 1];
for i in (0..n).rev() {
for j in 0..=i {
triangle[i][j] += triangle[i + 1][j].min(triangle[i + 1][j + 1]);
f[j] = f[j].min(f[j + 1]) + triangle[i][j];
}
}
triangle[0][0]
f[0]
}
}
```
Expand All @@ -284,28 +291,4 @@ impl Solution {

<!-- solution:end -->

<!-- solution:start -->

### 方法三

<!-- tabs:start -->

#### Python3

```python
class Solution:
def minimumTotal(self, triangle: List[List[int]]) -> int:
n = len(triangle)
for i in range(n - 2, -1, -1):
for j in range(i + 1):
triangle[i][j] = (
min(triangle[i + 1][j], triangle[i + 1][j + 1]) + triangle[i][j]
)
return triangle[0][0]
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- problem:end -->
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