defminPathSum(self, grid: List[List[int]]) ->int:"""dp[ij]表示从起点到ij最短距离,某位置只能来自左或上,求二者最小即可。 ij初始值为grid[ij],因此每次算dp[ij]时只要把左或上的dp值加过来即可"""m, n=len(grid), len(grid[0]) dp=gridforjinrange(1, n): dp[0][j] += dp[0][...
AI代码解释 publicclassSolution{publicintMinPathSum(int[][]grid){int m=grid.Length;int n=grid[0].Length;int[,]dp=newint[m+1,n+1];for(int i=2;i<=m;i++)dp[i,0]=int.MaxValue;for(int j=2;j<=n;j++)dp[0,j]=int.MaxValue;for(int i=1;i<=m;i++){for(int j=1;j<=n...
题目:A robot is located at the top-left corner of amxngrid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below). How many...
当i>0且 j>0时,dp[i][j]=min(dp[i−1][j],dp[i][j−1])+grid[i][j]。 最后得到 dp[m−1][n−1] 的值即为从网格左上角到网格右下角的最小路径和。 三、代码 代码语言:javascript 代码运行次数:0 运行 AI代码解释 class Solution { public int minPathSum(int[][] grid) { int...
return dp[h-1][w-1]; } }; class node { public: int x; int y; node(int x,int y) {this->x=x; this->y=y; } }; class Solution { public: int shortestPathBinaryMatrix(vector<vector<int>>& grid) { int ans = 0,path = 0; ...
; } }return dp[rows - 1][columns - 1]; }publicstaticvoidmain(String[] args){int[][] grid = newint[][]{{1, 3, 1}, {1, 5, 1}, {4, 2, 1}}; System.out.println(minPathSum(grid)); System.out.println(minPathSum2(grid)); }}【每日寄语】 穷人缺什么：...
class Solution { public: int minPathSum(vector<vector<int>>& grid) { if (grid.empty() || grid[0].empty()) return 0; int m = grid.size(), n = grid[0].size(); vector<vector<int>> dp(m, vector<int>(n)); dp[0][0] = grid[0][0]; for (int i = 1; i < m; ++i...
最后得到 dp[m−1][n−1] 的值即为从网格左上角到网格右下角的最小路径和。 三. 代码示例 class Solution(object): def minPathSum(self, grid): """ :type grid: List[List[int]] :rtype: int """ # 解法一, 空间优化 m, n = len(grid), len(grid[0]) ...
初始状态:dp[0][0] = grid[0][0] 代码实现: class Solution: def minPathSum(self, grid: List[List[int]]) -> int: # if not grid or not grid[0]: # return 0 m,n = len(grid),len(grid[0]) dp = [[0]*n for _ in range(m)] ...
接下来找状态转移方程,因为到达当前位置 (i, j) 只有两种情况,要么从上方 (i-1, j) 过来,要么从左边 (i, j-1) 过来,我们选择 dp 值较小的那个路径,即比较 dp[i-1][j] 和 dp[i][j-1],将其中的较小值加上当前的数字 grid[i][j],就是当前位置的 dp 值了...