[LeetCode] Unique Paths I, II

Unique Paths I

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.

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：Unique Paths I

Climbing Stairs二维版。计算解个数的题多半是用DP。而这两题状态也非常显然，dp[i][j]表示从起点到位置(i, j)的路径总数。DP题目定义好状态后，接下去有两个任务：找通项公式，以及确定计算的方向。

1. 由于只能向右和左走，所以对于(i, j)来说，只能从左边或上边的格子走下来：

dp[i][j] = dp[i-1][j] + dp[i][j-1]

2. 对于网格最上边和最左边，则只能从起点出发直线走到，dp[0][j] = dp[i][0] = 1

3. 计算方向从上到下，从左到右即可。可以用滚动数组实现。

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class Solution { public: int uniquePaths(int m, int n) { if(m<1 || n<1) return 0; vector<int> dp(n, 1); for(int i=1; i<m; i++) { for(int j=1; j<n; j++) { dp[j] += dp[j-1]; } } return dp[n-1]; } };

思路：Unique Paths II

与I有两点不同：

1. 当(i, j)有障碍时dp[i][j] = 0

2. dp[0][j]和dp[i][0]未必为1.

dp[0][j] = obstacleGrid[0][j] ? 0 : dp[0][j-1]

dp[i][0] = obstacleGrid[i][0] ? 0 : dp[i-1][0]

3. 当obstacleGrid [0][0] = 1时，return 0

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class Solution { public: int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) { if(obstacleGrid.empty() || obstacleGrid[0].empty() || obstacleGrid[0][0]==1) return 0; int m = obstacleGrid.size(), n = obstacleGrid[0].size(); vector<int> dp(n,1); for(int j=1; j<n; j++) { if(obstacleGrid[0][j]==1) dp[j] = 0; else dp[j] = dp[j-1]; } for(int i=1; i<m; i++) { dp[0] = obstacleGrid[i][0]==1 ? 0 : dp[0]; for(int j=1; j<n; j++) { dp[j] = obstacleGrid[i][j]==1 ? 0 : dp[j-1] + dp[j]; } } return dp[n-1]; } };