LeetCod: 64. Minimum Path Sum 最小路径和_leet minimum path sum-优快云博客

博客给出一道算法试题，要求在一个m x n的非负网格中，找出从左上角到右下角路径上数字和最小的路径，且只能右移或下移。还给出了解题思路，即一行一行统计所有最小和，并列举了四种情况。

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试题：
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.
代码：
由于只能往右或往下走，那么我们可以一行一行的统计出所有最小和。有四种情况：
1、行列都为0
2、行为0
3、行不为0，列为0
4、通用情况，从上，或者从左来的。

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