[]leetcode]Unique Paths II

更新时间：2022-09-17 15:15:18

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]
]

题目大意
寻求最短路径，从左上走到右下，保证每次只能往左走或往下走（不可以斜着走）。其中数字1是障碍，表示“此路不通”，求总共的路线数

思路
1. 如果没有障碍

val[i][0] = 1

val[0][j] = 1

val[i][j] = val[i-1][j] + val[i][j-1]

2. 有了障碍后

如果obstacle[i][j] = 1 
     val[i][j] = 1

否则

tmp = obstacle[i-1][j] == 1 ? 0 : val[i-1][j]

tmp = obstacle[i][j-1] == 1 ? tmp : tmp + val[i-1][j-1]

val[i][j] = tmp

参考代码

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
        int row = obstacleGrid.size();
        int col = obstacleGrid[0].size();
        int token = 1;
        int val[row][col];
        for (int j = 0; j < col; ++j)
        {
            if(obstacleGrid[0][j] == 1)
                token = 0;
            val[0][j] = token;
            
        }
        token = 1;
        for (int i = 0; i < row; ++i)
        {
            if(obstacleGrid[i][0] == 1)
                token = 0;
            val[i][0] = token;
        }
        for (int i = 1; i < row; ++i)
        {
            for(int j = 1; j < col; ++j)
            {
                if (obstacleGrid[i][j] == 1)
                    val[i][j] = 0;
                else
                {
                    int tmp = obstacleGrid[i-1][j] == 1 ? 0 :val[i-1][j];
                    tmp = obstacleGrid[i][j-1] == 1 ? tmp : tmp + val[i][j-1];
                    val[i][j] = tmp;
                }
            }
        }
        return val[row-1][col-1];
    }
};

本文转自jihite博客园博客，原文链接：http://www.cnblogs.com/kaituorensheng/p/3806896.html，如需转载请自行联系原作者

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[]leetcode]Unique Paths II

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