动态规划 最短路径和

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.


这类求最短最长的问题，实际的操作做法大多数都是定义一个数组，数组的下标代表对应问题，其值代表这种情况下的最优解最优解，然后答案的最优解就是对应下标数组的值。这里我们也创建一个二维数组dp[][],dp[i][j]代表从dp[0][0]
走到dp[i][j]的最短路径和。
dp[i][j] = Math.min(dp[i-1][j],dp[i][j-1])+grid[i][j];

public class MinPathSum{
    public int minPathSum(int grid[][]){
        final int m = grid.length;
        final int n = grid[0].length;
        int[][] dp = new int[m+1][n+1];
        
        dp[0][0] = grid[0][0];
        for(int i = 1; i < m; i++)
            dp[i][0] = dp[i-1][0] + grid[i][0];

        for(int j = 1; j < n; j++)
            dp[0][j] = dp[0][j-1] + grid[0][j];
        
        for(int i = 1; i < m; i++){
            for(int j = 1; j < n;j++)
                dp[i][j] = Math.min(dp[i-1][j],dp[i][j-1])+grid[i][j];

        return dp[m-1][n-1];
    }    
}

 

转载于:https://www.cnblogs.com/tiandiou/p/9695589.html