Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.
Note: You can only move either down or right at any point in time.
Example:
Input: [ [1,3,1], [1,5,1], [4,2,1] ] Output: 7 Explanation: Because the path 1→3→1→1→1 minimizes the sum.
Python version：
class Solution(object): def minPathSum(self, grid): """ :type grid: List[List[int]] :rtype: int """ m,n=len(grid),len(grid[0]) dp = [[2147483647] * (n+1) for _ in range(m+1)] dp[1][1]=grid[0][0] for i in range(1,m+1): for j in range(1,n+1): if i==j==1: continue dp[i][j]=min(dp[i-1][j],dp[i][j-1])+grid[i-1][j-1] #print(dp[-1][1]) return dp[m][n] solu=Solution() myList2=[ [1, 3, 1], [1, 5, 1], [4, 2, 1] ] print(solu.minPathSum(myList2))
C++ version:
#include<stdio.h> #include<iostream> #include<string> #include<vector> #include<set> #include<map> #include<algorithm> using namespace std; // Author: Huahua class Solution { public: int minPathSum(vector<vector<int>>& grid) { int m=grid.size(),n=grid[0].size(); vector<vector<int>> dp= vector<vector<int>>(m + 1, vector<int>(n + 1, 2147483647)); dp[1][1]=grid[0][0]; for(int i=1;i<m+1;i++) { for(int j=1;j<n+1;j++) { if(i==j&&j==1) { continue; } dp[i][j]=min(dp[i-1][j],dp[i][j-1])+grid[i-1][j-1]; } } return dp[m][n]; } }; int main() { Solution s; vector<vector<int>> grid= { {1,3,1}, {1,5,1}, {4,2,1} }; int res=s.minPathSum(grid); cout<<res<<endl; return 0; }
The difference is it want the min,so we initialize the array should use MAX_INT value 2147483647
We have to make dp larger than vector by +1,only in this way we can better solve the problem i-1 has minus one,if we want to initialize it to 0,it doesn't have matter for reason that it will be auto in 0,but we need it to be the MAX_INT we need the minimum value.So we have to use larger matrix but 63 needn't
70BONUS：
n is the input parameter
Two Attention Point:
1.because we initialize the 0-2 value so must let n+2 or the index may out of range.
2.dp[0] dp[1] this kinds of value need to initial by this way dp=[[0] for _ in range(N+2)]
3.first I let this position is 1000,but when I change into n+2 whole program quick a lot
4.range function can have two parameter (start,end) for i in range(3)---->0,1,2 for i in range(1,3)----->1,2
dp=[[0] for _ in range(n+2)] dp[0] = 0 dp[1] = 1 dp[2] = 2
