zoukankan      html  css  js  c++  java
  • 0063. Unique Paths II (M)

    Unique Paths II (M)

    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.

    Example 1:

    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
    

    题意

    在矩形中找到一条路径,起点为左上顶点,终点为右下顶点,路径中只能向右或向下走,且矩形中存在不能通过的障碍点,要求统计不同路径的个数。

    思路

    0062. Unique Paths (M) 方法一致,只是多了障碍点不好用组合数解决问题。使用动态规划,并用滚动数组进行优化。


    代码实现

    Java

    动态规划

    class Solution {
        public int uniquePathsWithObstacles(int[][] obstacleGrid) {
            int n = obstacleGrid.length, m = obstacleGrid[0].length;
            int[][] dp = new int[n][m];
            
            dp[0][0] = obstacleGrid[0][0] == 1 ? 0 : 1;
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    if (i != 0 || j != 0) {
                        dp[i][j] = obstacleGrid[i][j] == 1 ? 0 : 
                        		((i == 0 ? 0 : dp[i - 1][j]) + (j == 0 ? 0 : dp[i][j - 1]));
                    }
                }
            }
            
            return dp[n - 1][m - 1];
        }
    }
    

    滚动数组优化

    class Solution {
        public int uniquePathsWithObstacles(int[][] obstacleGrid) {
            int n = obstacleGrid.length, m = obstacleGrid[0].length;
            int[] dp = new int[m];
            
            dp[0] = 1;
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    dp[j] = obstacleGrid[i][j] == 1 ? 0 : j > 0 ? dp[j - 1] + dp[j] : dp[j];
                }
            }
            
            return dp[m - 1];
        }
    }
    

    JavaScript

    /**
     * @param {number[][]} obstacleGrid
     * @return {number}
     */
    var uniquePathsWithObstacles = function (obstacleGrid) {
      let n = obstacleGrid.length
      let m = obstacleGrid[0].length
      let scroll = new Array(m).fill(0)
      scroll[0] = 1
      for (let i = 0; i < n; i++) {
        for (let j = 0; j < m; j++) {
          if (obstacleGrid[i][j] === 1) {
            scroll[j] = 0
          } else {
            scroll[j] = j === 0 ? scroll[j] : scroll[j] + scroll[j - 1]
          }
        }
      }
      return scroll[m - 1]
    }
    
  • 相关阅读:
    巩固基础前台
    super 、static、final关键字加深记忆哦!还有父子类构造函数调用问题
    java异常了解
    spring粗略整体认识
    java枚举新认识
    泛型集合注意事项
    java反射基础
    对java集合类的认识——基础很重要
    多线程(C++ And POSIX)
    v2代理原理,应用
  • 原文地址:https://www.cnblogs.com/mapoos/p/13296884.html
Copyright © 2011-2022 走看看