Leetcode 题解系列（八）

64. Minimum Path Sum

题目要求

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.

Example 1:

[[1,3,1],
[1,5,1],
[4,2,1]]

Given the above grid map, return

7

. Because the path

1→3→1→1→1

minimizes the sum.

题目分析

暴力解法

一个格子可以从上面个格子走来，也可以从左边的格子走来，如果是N∗N地图，共有2N−1中选择。

动态规划

由于上一个点只能是上方或者左方的格子，那么到达该格子的最短路径即使从两个方向来的最短的一个。即是，distance[i][j]=min(from[i−1][j],from[i][j−1])

初始状态：起点的距离为0

实现如下：

class Solution {
public:
int minPathSum(vector<vector<int>> &grid) {
const auto size_r = grid.size(), size_c = grid[0].size();
auto map = vector<vector<int>>(size_r, vector<int>(size_c, 0));
for (int i = 1; i < size_c; ++i) {
map[0][i] = map[0][i - 1] + grid[0][i - 1];
}
for (int i = 1; i < size_r; ++i) {
map[i][0] = map[i - 1][0] + grid[i - 1][0];
for (int j = 1; j < size_c; ++j) {
map[i][j] = std::min(map[i - 1][j] + grid[i - 1][j],
map[i][j - 1] + grid[i][j - 1]);
}
}
// for (auto& i : map) {
//   for (auto& j : i) {
//     std::cout << j << " ";
//   }
//   std::cout << std::endl;
// }
return map[size_r - 1][size_c - 1] + grid[size_r - 1][size_c - 1];
}
};