[leetcode] 64. Minimum Path Sum

Question:

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.

Solution:

用动态规划，思路和做过的174. Dungeon Game是一样的,而且没那题那么复杂要考虑几种情况。
从最右下方开始往回遍历，状态转移方程为：

f(i, j) = min(f(i, j+1), f(i+1, j)) + nums(i, j)

即每次只可以去右边或者下边，选最小的那个走就行了。初始化时f(m, n) = nums(m, n) 并且注意边界的情况就好。

和174那题一样，空间上可以只用一维数组，可以用f(j+1) 代替 f(i, j+1) 因为f(j+1) 是最新更新的，用f(j) 代替f(i+1, j) 因为f(j) 是之前 (i-1) 那一层的。

class Solution {public:    int minPathSum(vector<vector<int>>& grid) {        if (grid.size() == 0)            return 0;        int m = grid.size(), n = grid[0].size();        vector<int> dp(n, 0);        dp[n-1] = grid[m-1][n-1];        for (int j = n-2; j >= 0; j--)            dp[j] = dp[j+1] + grid[m-1][j];        for (int i = m-2; i >= 0; i--) {            dp[n-1] += grid[i][n-1];            for (int j = n-2; j >= 0; j--) {                dp[j] = min(dp[j+1], dp[j]) + grid[i][j];            }        }        return dp[0];    }};