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# 120     Triangle

## 120. [Triangle](https://leetcode.com/problems/triangle/description/)

## 1. Question

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.

## 2. Implementation

**(1) DP**

思路: 根据题目三角形的特性，我们可以知道最底下的行的列数等于三角形的行数，dp\[i]\[j]表示在第i行和第j列的最小path sum, dp\[i]\[j] = triagnle.get(i).get(j) + Math.min(dp\[i + 1]\[j], dp\[i + 1]\[j + 1])

```java
class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        if (triangle == null || triangle.size() == 0) {
            return 0;
        }

        int n = triangle.size();
        int[][] dp = new int[n][n];

        for (int j = 0; j < triangle.get(n - 1).size(); j++) {
            dp[n - 1][j] = triangle.get(n - 1).get(j);
        }

        for (int i = n - 2; i >= 0; i--) {
            for (int j = 0; j < triangle.get(i).size(); j++) {
                dp[i][j] = triangle.get(i).get(j) + Math.min(dp[i + 1][j], dp[i + 1][j + 1]);
            }
        }
        return dp[0][0];
    }
}
```

## 3. Time & Space Complexity

**DP:** 时间复杂度O(n^2), 空间复杂度O(n^2)