intminPathSum(vector<vector<int> > &grid) { introw = grid.size(),col; if(row == 0)return0; elsecol = grid[0].size(); vector<int>dp(col+1, INT_MAX); dp[1] = 0; for(inti = 1; i <= row; i++) for(intj = 1; j <= col; j++) dp[j] = grid[i-1][j-1] + ...
Given amxngrid filled with non-negative numbers, find a path from top left to bottom right whichminimizesthe sum of all numbers along its path. Note: You can only move either down or right at any point in time. 该题解答参考自一博文。 设dp[i][j]表示从左上角到grid[i][j]的最小路径...
代码 funcminPathSum(grid[][]int)int{row:=len(grid)col:=len(grid[0])fori:=1;i<col;i++{grid[0][i]+=grid[0][i-1]}fori:=1;i<row;i++{grid[i][0]+=grid[i-1][0]}fori:=1;i<row;i++{forj:=1;j<col;j++{grid[i][j]+=min(grid[i-1][j],grid[i][j-1])}}ret...
Loading...空间复杂度O(m*n) class Solution { public int minPathSum(int[][] grid) { int m = grid.length; int n = grid[0].length; int[][] dp = new int[m][n]; dp[0][0] = grid[0][0]; for(int i = 1; i <…
public int minPathSum(int[][] grid) { int m=grid.length; int n=grid[0].length; if(m==0||n==0){ return 0; } int[][] minimumPathSum=new int[m][n]; minimumPathSum[0][0]=grid[0][0]; for (int i=1;i<m;i++){ ...
minimum-path-sum 因为限定了只能下和右,所以第一行和第一列边界就是直着走 class Solution { public: int minPathSum(vector<vector<int> > &grid) { int m=grid.size(),n=grid[0].size(); for(int i=0;i<m;i++) { for(int j=0;j<n;j++)...
这里我们直接用 grid 覆盖存,不去 new 一个 n 的空间了。 publicintminPathSum(int[][]grid){intm=grid.length;intn=grid[0].length;//由于第一行和第一列不能用我们的递推式,所以单独更新//更新第一行的权值for(inti=1;i<n;i++){grid[0][i]=grid[0][i-1]+grid[0][i];}//更新第一列...
intminPathSum(vector<vector<int>>&grid){ minSum=(~(unsignedint))>>; my_grid=grid; rowMax=grid.size(); colMax=grid[].size(); tra(,,); returnminSum; } }; 解法2:DP(还是不熟练,不太熟练递推dp和递归dp的区别,参考文章) dp[100][100];该dp数组记录的是每个位置上的最优解,即到达这一...
然后利用上面的两种途径找最短的作为到达grid[i][j]的最短路径。 代码: intminPathSum(vector<vector<int>>&grid){//C++introws=grid.size();if(rows==0)return0;intcols=grid[0].size();intupLine=0;for(inti=0;i<cols;i++)upLine+=grid[0][i];for(inti=1;i<rows;i++)upLine+=grid[i...
动规 // Minimum Path Sumgrid=grid;int[][](int=1;<m;i++){f[i][0]=f[i-1][0]+grid[i][0];}for(inti=1;i<n;i++){f[0][i