Leetcode 64. Minimum Path Sum 解题报告

---

题目

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.

解释
类似一个迷宫问题，从一个地图的左上角移动到右下角每次只能像下或者向上，每个格子中有一个数字，
问如何走才能使得走过的格子中的数字和最小。这是典型的BFS问题和动态规划，从起点开始扩散，
每次向2个方向扩散影响，同一个格子也之受自己左边的值和上方的值影响，所以可以得出状态转移方程：
cost[i][j] = min(cost[i-1][j]/表示上面的格子/, cost[i][j-1]/表示左边的格子/) + val[i][j]
其中cost表示走到每个格子需要的最小和，val表示每个格子中的数字。

---

AC代码

class Solution {    public int minPathSum(int[][] grid) {        int [][] dp = new int[grid.length][grid[0].length];        dp[0][0] = grid[0][0];        for (int i = 0; i < grid.length; ++i) {            for (int j = 0; j < grid[0].length; ++j) {                if (i == 0&&j==0) continue;                if (i==0&&j>0) dp[i][j] +=  ( dp[i][j-1] + grid[i][j]);                else if (j==0 && i > 0) dp[i][j] += ( dp[i-1][j] + grid[i][j]);                else dp[i][j] += ( Math.min(dp[i-1][j], dp[i][j-1]) + grid[i][j]);            }        }        return dp[grid.length-1][grid[0].length-1];    }}