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?

题目的意思是从Start到Finish,每次只能向下或右，有多少可能

自己最开始的想法：深度优先遍历即可（超时了）

void FindPort(int m,int n,int row,int col,int &count){    if (row == m &&col == n)    {         count++;    }    else if (row <= m&&col <= n)    {        for (int i = 0; i < 2; i++)        {            if (i == 0) //向右            {                col++;                FindPort(m, n, row, col, count);                col--;            }            else if (i==1)//向下            {                row++;                FindPort(m, n, row, col, count);                row--;            }        }    }}int uniquePaths(int m, int n){    int count = 0;    FindPort(m, n, 1, 1, count);    return count;}

解法：DP 很简单

int uniquePaths(int m, int n){     vector<vector<int> > path(m, vector<int> (n, 1));        for (int i = 1; i < m; i++)            for (int j = 1; j < n; j++)                path[i][j] = path[i - 1][j] + path[i][j - 1];//上一个位置可能是左边，也可能是右边        return path[m - 1][n - 1];}

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.

在第一题中加了一个障碍物，有障碍物的不能到达

int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid){    int m = obstacleGrid.size();    int n = obstacleGrid[0].size();    vector<vector<int> > path(m, vector<int>(n, 0));    for (int k = 0; k < n; k++)    {        if (obstacleGrid[0][k] == 0)            path[0][k] = 1;        else break; //若出现障碍物，则第一行的后面几列无法到达    }    for (int h = 0; h < m; h++)    {        if (obstacleGrid[h][0] == 0)            path[h][0] = 1;        else break;//若出现障碍物，则第一列的后面几行无法到达    }    if (obstacleGrid[m - 1][n - 1] == 1) return 0;    for (int i = 1; i < m; i++)    for (int j = 1; j < n; j++)    {        if (obstacleGrid[i - 1][j] == 0 && obstacleGrid[i][j - 1] == 0)            path[i][j] = path[i - 1][j] + path[i][j - 1];        else if (obstacleGrid[i - 1][j] == 0)            path[i][j] = path[i - 1][j];        else if (obstacleGrid[i][j - 1] == 0)            path[i][j] = path[i][j - 1];    }    return path[m - 1][n - 1];}