本文共 1772 字，大约阅读时间需要 5 分钟。
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 > &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