动态规划——leetcode刷题_OJ 62&64_动态分配ip地址 算法 oj

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).

How many possible unique paths are there?

Above is a 7 x 3 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

Example 1:

Input: m = 3, n = 2
Output: 3
Explanation:
From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
1. Right -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right
Example 2:

Input: m = 7, n = 3
Output: 28
题目的意思：给出m,n表示m*n的格子，求从左上角走到右下角共有几条路线，注意只能向右或者向下走。

动规我真的学得不好，所以从一些经典的题目开始做起，再练一下基础。

class Solution {
    public int uniquePaths(int m, int n) {//m列,n行：dp[x][y]=dp[x-1][y]+dp[x][y-1]
        int[][] board=new int[m][n];
        for(int i=0;i<m;i++){
            for (int j=0;j<n; j++) {
                if(i ==0 || j == 0){
                    board[i][j]=1;
                }
                else board[i][j]=board[i-1][j]+board[i][j-1];
            }
        }
        return board[m-1][n-1];
    }
}

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.

题目的意思：给出一个二维矩阵，每一点有权重，找出从左下角走到右上角的权重最小的路径的权重和。

一开始我看着这像一道要用地杰斯特拉算法的求最短路径的题目，可是Dis算法需要自己构建图，而这里已经有一个很规律的二维矩阵了，所以跟上面那题很类似，动规修改原输入数据后叠加就行。

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