算法精粹(algorithm-essentials)

感谢soulmachine@github提供内容
## Triangle


### 描述

Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.

For example, given the following triangle

```
[
[2],
[3,4],
[6,5,7],
[4,1,8,3]
]
```

The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11).

Note: Bonus point if you are able to do this using only `O(n)` extra space, where n is the total number of rows in the triangle.


### 分析

设状态为`f(i, j)`,表示从从位置`(i,j)`出发,路径的最小和,则状态转移方程为

$$
f(i,j)=\min\left\{f(i+1,j),f(i+1,j+1)\right\}+(i,j)
$$


### 代码

{% if book.java %}
```java
// Triangle
// 时间复杂度O(n^2),空间复杂度O(1)
public class Solution {
public int minimumTotal(List> triangle) {
for (int i = triangle.size() - 2; i >= 0; --i)
for (int j = 0; j < i + 1; ++j) {
int old = triangle.get(i).get(j);
triangle.get(i).set(j, old + Math.min(triangle.get(i + 1).get(j),
triangle.get(i + 1).get(j + 1)));
}

return triangle.get(0).get(0);
}
}
```
{% endif %}

{% if book.cpp %}
```cpp
// Triangle
// 时间复杂度O(n^2),空间复杂度O(1)
class Solution {
public:
int minimumTotal (vector>& triangle) {
for (int i = triangle.size() - 2; i >= 0; --i)
for (int j = 0; j < i + 1; ++j)
triangle[i][j] += min(triangle[i + 1][j],
triangle[i + 1][j + 1]);

return triangle [0][0];
}
};
```
{% endif %}