1: class Solution {
2: public:
3: int minPathSum(vector<vector<int>>& grid) {
4: int row = grid.size();
5: if (row == 0) return 0;
6: int col = grid[0].size();
7: for (int j = 1; j < col; j++) {
8: grid[0][j] += grid[0][j-1];
9: }
10: for (int i = 1; i < row; i++) {
11: grid[i][0] += grid[i-1][0];
12: }
13: for (int i = 1; i < row; i++) {
14: for (int j = 1; j < col; j++) {
15: grid[i][j] += min(grid[i-1][j], grid[i][j-1]);
16: }
17: }
18: return grid[row-1][col-1];
19: }
20: };
Sunday, August 14, 2016
64. Minimum Path Sum
Let dp[i][j] to be the minimum path sum at (i, j), then the transition state is dp[i][j] = grid[i][j] + min(dp[i-1][j], dp[i][j-1]).
Labels:
dynamic programming,
leetcode,
matrix
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