Unique Paths I&&II

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.

方法一：递归 f(m,n) = f(m-1, n) + f(m, n-1)，m=1或n=1时，f(m,n) = 1

class Solution {
public:
int uniquePaths(int m, int n) {
if( m==1 || n==1 ) return 1;
return uniquePaths(m-1,n) + uniquePaths(m, n-1);
}
};

方法二：使用dp求解，dp[i][j] = dp[i-1][j] + dp[i][j-1], i = 1或j=1时，dp[i][j] = 1，代码可优化，便于理解不化~~

class Solution {
public:
int uniquePaths(int m, int n) {
vector< vector<int> > dp( m+1, vector<int>(n+1, 0) );
for(int i=1; i<=m; ++i) dp[i][1] = 1;
for(int i=1; i<=n; ++i) dp[1][i] = 1;
for(int i=2; i<=m; ++i)
for(int j=2; j<=n; ++j)
dp[i][j] = dp[i-1][j] + dp[i][j-1];
return dp[m]
;
}
};

方法三：利用组合公式，格子走位问题，其实就是向下走m-1步，向右走n-1步，总共走m+n-2步，即在m+n-2步中挑选m-1步向下走，其余的步向右走

class Solution {
public:
int uniquePaths(int m, int n) {
return combination(m+n-2, m-1);
}

int combination(int n, int m) {
if( m > (n>>1) ) m = n-m;
long long ans = 1;
for(int i=1; i<=m; ++i)
ans = ans * (n-i+1) / i;　　//不能写成ans *= (n-i+1)/i;
return ans;
}
};

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.

利用上题方法二，如果obstacle[i][j] = true，那么dp[i][j] = 0 否则 dp[i][j] = dp[i-1][j] + dp[i][j-1]

class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
int m = obstacleGrid.size();
int n = obstacleGrid[0].size();
if( m < 1 || n < 1 ) return 0;
vector< vector<int> > dp(m, vector<int>(n, 0));
for(int i=0; i<n; ++i)  //初始化很重要，若碰到一个障碍，那后面的都不能到达
if( !obstacleGrid[0][i] ) dp[0][i] = 1;
else break;
for(int i=0; i<m; ++i)
if( !obstacleGrid[i][0] ) dp[i][0] = 1;
else break;
for(int i=1; i<m; ++i)
for(int j=1; j<n; ++j)
if( !obstacleGrid[i][j] ) dp[i][j] = dp[i-1][j] + dp[i][j-1];
return dp[m-1][n-1];
}
};