刚开始思考用递归来解决该问题:
class Solution { public: int minSum(vector<vector<int> > &grid, int m, int n,int i, int j){ if(i == m && j == n){ return grid[i][j]; }else if(i == m){ return grid[i][j] + minSum(grid,m,n,i,j+1); }else if(j == n){ return grid[i][j] + minSum(grid,m,n,i+1,j); } int right = grid[i][j] + minSum(grid,m,n,i,j+1); int down = grid[i][j] + minSum(grid,m,n,i+1,j); int min = right < down ? right : down; return min; } int minPathSum(vector<vector<int> >& grid) { int m = grid.size()-1; int n = grid[0].size()-1; return minSum(grid,m,n,0,0); } };
结果运行结果正确,但是超时了
于是参考讨论区别人的做法有更快的解决方案,O(m*n)的时间复杂度,依次求出到每个点的最小路径和,一直到最右下角的格子,算法成功
class Solution { public: int minPathSum(vector<vector<int>>& grid) { if(grid.size() == 0) return 0; int m = grid.size(), n = grid[0].size(); for(int i=1;i<m;i++) grid[i][0] += grid[i-1][0]; for(int i=1;i<n;i++) grid[0][i] += grid[0][i-1]; for(int i=1;i<m;i++) for(int j=1;j<n;j++) grid[i][j] += min(grid[i-1][j], grid[i][j-1]); return grid[m-1][n-1]; } };