分类:algorithm| 发布时间:2017-03-17 14:08:00
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.
m x n 的格子填充了非负整数,找出一个从左上到右下的路径使得所经过的格子的和最小。
class Solution {
public:
int minPathSum(vector<vector<int>>& grid) {
int m = grid.size(), n = grid[0].size();
vector<vector<int>> sums(m, vector<int>(n, 0));
sums[0][0] = grid[0][0];
// init first col
for (int i = 1; i < m; ++i) {
sums[i][0] = grid[i][0] + sums[i - 1][0];
}
// init first row
for (int i = 1; i < n; ++i) {
sums[0][i] = grid[0][i] + sums[0][i - 1];
}
for (int i = 1; i < m; ++i) {
for (int j = 1; j < n; ++j) {
sums[i][j] = min(sums[i - 1][j], sums[i][j - 1]) + grid[i][j];
}
}
return sums[m - 1][n - 1];
}
};