迷宫里的动态规划应用

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[LeetCode 63] Unique Paths II

题目

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).
Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as 1 and 0 respectively in the grid.
Note: m and n will be at most 100.

测试案例

Input:
[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]
Output: 2
Explanation:
There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:
1. Right -> Right -> Down -> Down
2. Down -> Down -> Right -> Right

思路

记 nums[i][j] 表示从 (i,j) 点到达右下角的不同路径数。那么有如下递归式：
\[ nums[i][j] = \lbrace \begin{align} nums[i + 1][j] + num[i][j + 1] , \;(i,j) 处无障碍 \0,\;(i,j) 处有障碍 \end{align} \]

代码如下

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m, n;        
        if((m = obstacleGrid.length) < 1 || (n = obstacleGrid[0].length) < 1){
            return 0;
        }                
        int[][] nums = new int[m][n];
        if((nums[m - 1][n - 1] = 1 - obstacleGrid[m - 1][n - 1]) == 0){
            return 0;
        }
        for(int i = n - 2; i > -1; i--){
            if(obstacleGrid[m - 1][i] == 1){
                nums[m - 1][i] = 0;
            }
            else{
                nums[m - 1][i] = nums[m - 1][i + 1];
            }            
        }
        for(int i = m - 2; i > -1; i--){
            if(obstacleGrid[i][n - 1] == 1){
                nums[i][n - 1] = 0;
            }
            else{
                nums[i][n - 1] = nums[i + 1][n - 1];    
            }            
        }
        for(int i = m - 2; i > -1; i--){
            for(int j = n - 2; j > -1; j--){
                if(obstacleGrid[i][j] == 1){
                    nums[i][j] = 0;
                }
                else{
                    nums[i][j] = nums[i + 1][j] + nums[i][j + 1];
                }
            }
        }
        return nums[0][0];
    }
}

[LeetCode 64] Minimum Path Sum

题目

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.

测试案例

Input:
[
  [1,3,1],
  [1,5,1],
  [4,2,1]
]
Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.

代码如下

class Solution {
    public int minPathSum(int[][] grid) {
        int m = grid.length, n = grid[0].length;        
        int[][] price = new int[m][n];
        price[m - 1][n - 1] = grid[m - 1][n - 1];
        for(int i = n -2; i > -1; i--){
            price[m - 1][i] = price[m - 1][i + 1] + grid[m - 1][i];
        }
        for(int i = m - 2; i > -1; i--){
            price[i][n - 1] = price[i + 1][n - 1] + grid[i][n - 1];
        }
        for(int i = m - 2; i> -1; i--){
            for(int j = n -2; j > -1; j--){
                price[i][j] = price[i + 1][j];
                if(price[i][j] > price[i][j + 1]){
                    price[i][j] = price[i][j + 1];
                }
                price[i][j] += grid[i][j];
            }
        }
        return price[0][0];
    }
}

迷宫里的动态规划应用

标签：one   ace   minimum   bottom   use   ber   start   tput   find   

原文地址：https://www.cnblogs.com/echie/p/9594499.html