Monday, November 24, 2014

[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];
    }
};

3 comments:

  1. unique path I 其实是一个 n选k的问题. 总共要往右走m-1步,往下走n-1步,所以是m+n-1里面选择n-1个.这个计算需要loop x次,x为max(m,n).所以最快可以是linear.

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    Replies
    1. 是C(m + n - 2, m - 1) or C(m + n - 2, n - 1)吧。虽然用组合数学很好得出这个,但是计算C(..., ...)的结果的时间复杂度可能要高于DP了。

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