【Leetcode】Minimum Path Sum-优快云博客

本文链接：https://blog.youkuaiyun.com/yyz20002008/article/details/39753603

本文详细介绍了解决最小路径和问题的两种动态规划方法：二维DP和一维DP。通过实例演示了如何使用这两种算法从网格的左上角到右下角找到最小路径和。

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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.

Java:

1.二维DP

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

2.一维DP

http://blog.youkuaiyun.com/linhuanmars/article/details/22257673

递推式是res[i][j]=min(res[i-1][j], res[i][j-1])+grid[i][j]。总共进行两层循环，时间复杂度是O(m*n)。而空间复杂度仍是使用Unique Paths中的方法来省去一维，是O(m)，不了解的朋友可以看看哈。代码如下：

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