Leetcode-120Triangle

120. Triangle

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).

My Solution(C/C++)

#include 
#include 
#include 
#include 

using namespace std;

class Solution {
public:
    //vector> minimumTotal(vector> &triangle) {
    int minimumTotal(vector> &triangle) {
        int row = triangle.size();
        vector> dp(row, vector());
        for (int i = row - 1; i >= 0; i--) {
            for (int j = 0; j < triangle[i].size(); j++) {
                if (i == row - 1) {
                    dp[i].push_back(triangle[i][j]);
                }
                else {
                    dp[i].push_back(triangle[i][j] + min(dp[i + 1][j], dp[i + 1][j + 1]));
                }
            }
        }
        //return dp;
        return dp[0][0];
    }
};

int main() {
    vector> triangle(4, vector());
    triangle[0].push_back(2);
    triangle[1].push_back(3);
    triangle[1].push_back(4);
    triangle[2].push_back(6);
    triangle[2].push_back(5);
    triangle[2].push_back(7);
    triangle[3].push_back(4);
    triangle[3].push_back(1);
    triangle[3].push_back(8);
    triangle[3].push_back(3);
    //printf("%d ", triangle[1][1]);
    Solution s;
    //vector> dp;
    //dp = s.minimumTotal(triangle);
    //for (int i = 0; i < dp.size(); i++) {
    //  for (int j = 0; j < dp[i].size(); j++) {
    //      printf("%d ", dp[i][j]);
    //  }
    //  printf("\n");
    //}
    printf("三角形最小路径和：%d", s.minimumTotal(triangle));
    return 0;
}

My Solution(Python)

class Solution:
    def minimumTotal(self, triangle):
        """
        :type triangle: List[List[int]]
        :rtype: int
        """
        row = len(triangle) - 1
        column = len(triangle[len(triangle)-1]) - 1
        dp = [[] for i in range(row + 1)]
        for i in range(column + 1):
            dp[row].append(triangle[row][i])
        for r in range(row - 1, -1, -1):
            for c in range(len(triangle[r])):
                dp[r].append(min(dp[r+1][c], dp[r+1][c+1]) + triangle[r][c])
        return dp[0][0]