leetcode 之 Unique Paths

Unique Paths

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?

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

Note: m and n will be at most 100.

思路：典型的动态规划，dp[i][j]表示到matrix[i][j]的路径个数。则dp[i][j] = dp[i-1][j] + dp[i][j-1]。

int uniquePaths(int m, int n) { if(m <=0 || n <= 0)return 0; vector<vector<int> > dp(m); int i,j; for(i=0;i<m;i++) { vector<int> tmp(n,1);//至少一条 dp[i] = tmp; } for(i = 1;i < m;i++) { for(j = 1;j < n;j++) { dp[i][j] = dp[i-1][j] + dp[i][j-1]; } } return dp[m-1][n-1]; }

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.

思路：和上面类似，仅仅是当obstacleGrid[i][j] == 0时要把dp[i][j]=0。表示此路不通，初始化时也要注意

class Solution { public: int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) { int rows = obstacleGrid.size(); if(rows <= 0)return 0; int cols = obstacleGrid[0].size(); if(cols <= 0)return 0; vector<vector<int> > dp(rows); int i,j; for(i = 0;i < rows;i++) { vector<int> tmp(cols); dp[i] = tmp; } dp[0][0] = obstacleGrid[0][0] == 0 ? 1 : 0; for(i = 1;i < rows;i++)dp[i][0] = obstacleGrid[i][0] == 0 ? dp[i-1][0] : 0;//当为0时，不能简单的初始化为1。要初始化为前面的值，由于可能被前面挡住了 for(j = 1;j < cols;j++)dp[0][j] = obstacleGrid[0][j] == 0 ?

dp[0][j-1] : 0; for(i = 1;i < rows;i++) { for(j = 1;j < cols;j++) { dp[i][j] = obstacleGrid[i][j] == 0 ? dp[i-1][j] + dp[i][j-1] : 0; } } return dp[rows-1][cols-1]; } };

 

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转载于:https://www.cnblogs.com/bhlsheji/p/4757258.html

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