共计 803 个字符，预计需要花费 3 分钟才能阅读完成。
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:
Input:
[
[1,3,1],
[1,5,1],
[4,2,1]
]
Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.
难度：medium
题目：给定 m * n 的格子，找出从左上角到右下角和最小的路径。
思路：动态规划，状态转移方程 grid[i][j] = grid[i][j] + min(grid[i – 1][j], grid[i][j – 1]
Runtime: 4 ms, faster than 97.83% of Java online submissions for Minimum Path Sum.Memory Usage: 40.8 MB, less than 1.47% of Java online submissions for Minimum Path Sum.
class Solution {
public int minPathSum(int[][] grid) {
int m = grid.length, n = grid[0].length;
for (int i = 1; i < n; i++) {
grid[0][i] += grid[0][i – 1];
}
for (int i = 1; i < m; i++) {
grid[i][0] += grid[i – 1][0];
}
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
grid[i][j] += Math.min(grid[i – 1][j], grid[i][j – 1]);
}
}
return grid[m – 1][n – 1];
}
}
