• Unique Paths II


    Follow up for "Unique Paths":

    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.

    class Solution {
    public:
        int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) 
        {
            int m=obstacleGrid.size();
            if(m==0return 0;
            
            int n=obstacleGrid[0].size();
            int path[n];
            path[0]=1-obstacleGrid[0][0];
            for(int i=1;i<n;i++) 
                if(obstacleGrid[0][i]==1) path[i]=0;
                else path[i]=path[i-1];
            
            for(int i=1;i<m;i++)
            {
                if(obstacleGrid[i][0]==1) path[0]=0;
                for(int j=1;j<n;j++)
                    if(obstacleGrid[i][j]==1) path[j]=0;
                    else path[j]=path[j-1]+path[j];
            }
            return path[n-1];
        }
    }; 
  • 相关阅读:
    FFT-C语言
    C语言解析WAV音频文件
    图基(Tukey)检验
    方差分析中均值比较的方法
    模拟信号采样过程
    FS,FT,DFS,DTFT,DFT,FFT的联系和区别
    枚举enum类型
    样本概率统计
    宏和内联函数
    变量的生存期和存储分配
  • 原文地址:https://www.cnblogs.com/erictanghu/p/3759458.html
Copyright © 2020-2023  润新知