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.

算法1:最容易想到的是递归解法,uniquePaths(m, n) = uniquePaths(m, n-1) + uniquePaths(m-1, n), 递归结束条件是m或n等于1,这个方法oj超时了

 class Solution {
public:
int uniquePaths(int m, int n) {
if(m == || n == )return ;
else return uniquePaths(m, n - ) + uniquePaths(m - , n);
}
};

算法2:动态规划,算法1的递归解法中,其实我们计算了很多重复的子问题,比如计算uniquePaths(4, 5) 和 uniquePaths(5, 3)时都要计算子问题uniquePaths(3, 2),再者由于uniquePaths(m, n) = uniquePaths(n, m),这也使得许多子问题被重复计算了。要保存子问题的状态,这样很自然的就想到了动态规划方法,设dp[i][j] = uniquePaths(i, j), 那么动态规划方程为:

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

上述过程其实是从左上角开始,逐行计算到达每个格子的路线数目,由递推公式可以看出,到达当前格子的路线数目和两个格子有关:1、上一行同列格子的路线数目;2、同一行上一列格子的路线数目。据此我们可以优化上面动态规划方法的空间:

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

算法3:其实这个和组合数有关,对于m*n的网格,从左上角走到右下角,总共需要走m+n-2步,其中必定有m-1步是朝右走,n-1步是朝下走,那么这个问题的答案就是组合数:, 这里需要注意的是求组合数时防止乘法溢出        本文地址

 class Solution {
public:
int uniquePaths(int m, int n) {
return combination(m+n-, m-);
} int combination(int a, int b)
{
if(b > (a >> ))b = a - b;
long long res = ;
for(int i = ; i <= b; i++)
res = res * (a - i + ) / i;
return res;
}
};

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.

这一题可以完全采用和上一题一样的解法,只是需要注意dp的初始化值,和循环的起始值

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

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