【LeetCode】64. Minimum Path Sum（C++）

地址：https://leetcode.com/problems/minimum-path-sum/

题目：

Given a $m * n$ 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:

理解：

和上面的题一样，只是有了权值，加一下就可以了。

实现：

下面的实现很慢 跑了12ms 不太理解为什么。。

class Solution {
public:
	int minPathSum(vector<vector<int>>& grid) {
		for (int i = 1; i < grid.size(); ++i)
			grid[i][0] += grid[i - 1][0];
		for (int j = 1; j < grid[0].size(); ++j)
			grid[0][j] += grid[0][j - 1];
		for (int i = 1; i < grid.size(); ++i)
			for (int j = 1; j < grid[0].size(); ++j)
				grid[i][j] = min(grid[i][j] + grid[i - 1][j],grid[i][j] + grid[i][j - 1]);
		return grid.back().back();
	}
};

优化一下

class Solution {
public:
	int minPathSum(vector<vector<int>>& grid) {
		int r = grid.size(), c = grid[0].size();
		int **dp = new int*[r];
		for (int i = 0; i < r; ++i)
			dp[i] = new int[c];
		dp[0][0] = grid[0][0];
		for (int i = 1; i < r; ++i)
			dp[i][0] = dp[i - 1][0] + grid[i][0];
		for (int j = 1; j < c; ++j)
			dp[0][j] = dp[0][j - 1] + grid[0][j];
		for (int i = 1; i < r; ++i)
			for (int j = 1; j < c; ++j)
				dp[i][j] = grid[i][j] + min(dp[i - 1][j],dp[i][j - 1]);
		int res = dp[r - 1][c - 1];
		for (int i = 0; i < r; ++i)
			delete[]dp[i];
		delete[] dp;
		return res;
	}
};