三角形中最小路径和(leetcode120)

一：解题思路

方法一：动态规划的方法。定义d(i,j)为坐标(i,j)的最小路径和。状态转移方程为：d(i,j)=min(d(i-1,j-1),d(i-1,j))+a(i,j)

d(0,0)=a(0,0)，d(i,0)=d(i-1,0)+a(i,0)，d(i,i)=d(i-1,i-1)+a(i,i)

二：完整代码示例 (C、C++、Java、Python)

方法一C++:

class Solution {
public:
    int minimumTotal(vector<vector<int>>& triangle) {
        int n = triangle.size();
        vector<vector<int>> d(n,vector<int>(n,0));
        d[0][0] = triangle[0][0];

        for (int i = 1; i < n; i++) {
            d[i][0] = d[i - 1][0] + triangle[i][0];
            d[i][i] = d[i - 1][i - 1] + triangle[i][i];
            
            for (int j = 1; j < i; j++) {
                d[i][j] = min(d[i - 1][j - 1], d[i - 1][j]) + triangle[i][j];
            }
        }
        int minValue = d[n-1][0];
        for (int j = 1; j < n; j++) {
            minValue = min(minValue, d[n-1][j]);
        }

        return minValue;
    }
};

 

 

 

方法一Java:

class Solution {
      public int minimumTotal(List<List<Integer>> triangle) {
            int n=triangle.size();
            int[][] d=new int[n][n];
            d[0][0]=triangle.get(0).get(0);

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

 

 

方法一Python:

class Solution:
    def minimumTotal(self, triangle: List[List[int]]) -> int:
        n=len(triangle)
        d=[[0]*n for i in range(n)]
        d[0][0]=triangle[0][0]

        for i in range(1,n):
            d[i][0]=d[i-1][0]+triangle[i][0]
            d[i][i]=d[i-1][i-1]+triangle[i][i]
            for j in range(1,i):
                d[i][j]=min(d[i-1][j-1],d[i-1][j])+triangle[i][j]
        minValue=d[n-1][0]
        for j in range(1,n):
            minValue=min(minValue,d[n-1][j])
        
        return minValue