LeetCode-Minimum Path Sum[dp]

Minimum Path Sum

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.

分析：

动态规划的经典题目，设dp[i][j]表示从位置[0,0]到[i,j]的最小路径和，要到达位置[i,j]只能从[i,j-1]或[i-1,j]向右或向下走一步到达，

所以状态转移方程为：

dp[i][j]=min(dp[i-1][j],dp[i][j-1])+grid[i][j];

由二维dp进一步优化为一维dp：

当j=0时，dp[j]=dp[j]+grid[i][j];

当0<j&&j<n时，dp[j]=min(dp[j],dp[j-1])+grid[i][j];

(此时的dp[j]等价于二维dp中的dp[i][j])

参考代码：

public class Solution {
    public int minPathSum(int[][] grid) {
        int nlen=grid.length;
        int mlen=grid[0].length;
        int dp[]=new int[mlen];
        dp[0]=grid[0][0];
        for(int j=1;j<mlen;j++){
            dp[j]=dp[j-1]+grid[0][j];
        }
        for(int i=1;i<nlen;i++){
            for(int j=0;j<mlen;j++){
                dp[j]=grid[i][j]+(j==0?(dp[j]):(Math.min(dp[j-1], dp[j])));
            }
        }
        return dp[mlen-1];
    }
}