zoukankan      html  css  js  c++  java
  • 【题解】【矩阵】【回溯】【Leetcode】Unique Paths II

    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?

    NewImage

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

    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.

    For example,

    There is one obstacle in the middle of a 3x3 grid as illustrated below.

    [
      [0,0,0],
      [0,1,0],
      [0,0,0]
    ]
    

    The total number of unique paths is 2.

    Note: m and n will be at most 100.

    思路:

    很典型的回溯问题,到[i,j]的路径等于到[i-1,j]和[i,j-1]的路径之和,然后为了避免重复求解子问题,结合查表法记录子问题结果。编码要点在于处理 trivial case。

    代码:

     1 int uniquePaths2Node(vector<vector<int> > &oGrid, vector<vector<int> > &paths, int i, int j){
     2     //if(i < 0 || j < 0) return 0;//failed [[1]]=>0, 应该把控制条件放下面,不给调用不valid的子问题
     3     if(oGrid[i][j] == 1) return 0;
     4     
     5     if(i == 0 && j == 0) return 1 ^ oGrid[0][0];//failed [[0]]=>1
     6 
     7     int P = 0;
     8     if(i > 0 && oGrid[i-1][j] != 1){
     9         if(paths[i-1][j] == -1)
    10             paths[i-1][j] = uniquePaths2Node(oGrid, paths, i-1, j);
    11         P += paths[i-1][j];
    12     }
    13     if(j > 0 && oGrid[i][j-1] != 1){
    14         if(paths[i][j-1] == -1)
    15             paths[i][j-1] = uniquePaths2Node(oGrid, paths, i, j-1);
    16         P += paths[i][j-1];
    17     }
    18     return P;
    19 }
    20 int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
    21     int m = obstacleGrid.size();
    22     if(m == 0) return 0;
    23     int n = obstacleGrid[0].size();
    24     if(n == 0) return 0;
    25 
    26     vector<vector<int> > paths(m, vector<int>(n, -1));
    27     return uniquePaths2Node(obstacleGrid, paths, m-1, n-1);
    28 }
  • 相关阅读:
    开发微博应用7构思草图
    微博应用研究【2】
    跟着Artech学习WCF扩展(4) 扩展MessageInspector
    ASP.NET开源MVC框架VICI 测试的便利性
    第一次踏出.net后花园(一)
    回忆被三层架构忽悠的日子,上当的同学自觉举手
    微博应用开发10
    开发微博应用【5】应用的使用频率
    微博应用研究(4)
    微博应用研究(3)
  • 原文地址:https://www.cnblogs.com/wei-li/p/UniquePathsII.html
Copyright © 2011-2022 走看看