120. Triangle

短命女 2023-07-03 04:58 9阅读 0赞

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.

题意：给定三角形，求从顶部到底部的最短距离。

方法一：动态规划

转移方程为Dp(i, j) = min(Dp(i + 1, j) + Dp(i+1,  j +1)) + triangle\[i\]\[j\] ;  Dp(i, j)表示从最底层开始，走到Dp(i,j)路径和的最小值。

class Solution {
    public:
        int minimumTotal(vector<vector<int>>& triangle) {
            vector<vector<int>> result;
            result= triangle;
    
            for (int i = triangle.size() - 2; i >= 0; i--)
            {
                for (int j = 0; j < triangle[i].size(); j++)
                {
                    result[i][j] = min(result[i + 1][j], result[i + 1][j + 1]) + triangle[i][j];
                }
            }
            return result[0][0];
        }
    };

方法二：动态规划 + 状态压缩。

动态规划的思路同上。

复用记录状态的数组，使用一维数组，减少内存消耗。

class Solution {
    public:
        int minimumTotal(vector<vector<int>>& triangle) {
            vector<int> result;
            result= triangle[triangle.size()-1];
    
            for (int i = triangle.size() - 2; i >= 0; i--)
            {
                for (int j = 0; j < triangle[i].size(); j++)
                {
                    result[j] = min(result[j], result[j + 1]) + triangle[i][j];
                }
            }
            return result[0];
        }
    };