LeetCode 62. Unique Paths
2017-07-06 13:08
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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?
![](https://leetcode.com/static/images/problemset/robot_maze.png)
Above is a 3 x 7 grid. How many possible unique paths are there?
Note: m and n will be at most 100.
题目要求找出一个m*n网格中从左上角走到右下角的路径的数量,其中,每一步都只能向下或者向右走。在这种情况下,我想到了两种方式。
(1)设函数f(i,j)为从左上角走至第i行第j列的路径总数,显然,当i或者j等于1时,结果为1,而i,j都不等于1时,我们分析可知,到达格子(i,j)只有两种途径,一是从(i,j-1)往右移,二是从(i-1,j)往下移。所以,f(i,j)= f(i-1,j)+ f(i,j-1)。因此,可通过这个方法求出f(i,j)的值。
class Solution {
public:
int uniquePaths(int m, int n) {
int res[m+1][n+1];
for(int i=1;i<=m;i++)
res[i][1]=1;
for(int i=1;i<=n;i++)
res[1][i]=1;
for(int i=2;i<=m;i++){
for(int j=2;j<=n;j++){
res[i][j]=res[i-1][j]+res[i][j-1];
}
}
return res[m]
;
}
};
(2)单纯从移动来考虑,由于只能向右或者向下移动,所以从(1,1)移动到(m,n)共需要m+n-2步,其中向下m-1步,向右n-1步。所以说,在这m+n-2步中,需要考虑的就是何时向下何时向右,这就变成了排列组合的问题。
class Solution {
public:
int uniquePaths(int m, int n) {
int a=m+n-2,b=max(m,n)-1;
long s=1,t=1;
for(int i=b+1;i<=a;i++)
s*=i;
for(int i=2;i<=a-b;i++)
t*=i;
return s/t;
}
};
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?
![](https://leetcode.com/static/images/problemset/robot_maze.png)
Above is a 3 x 7 grid. How many possible unique paths are there?
Note: m and n will be at most 100.
题目要求找出一个m*n网格中从左上角走到右下角的路径的数量,其中,每一步都只能向下或者向右走。在这种情况下,我想到了两种方式。
(1)设函数f(i,j)为从左上角走至第i行第j列的路径总数,显然,当i或者j等于1时,结果为1,而i,j都不等于1时,我们分析可知,到达格子(i,j)只有两种途径,一是从(i,j-1)往右移,二是从(i-1,j)往下移。所以,f(i,j)= f(i-1,j)+ f(i,j-1)。因此,可通过这个方法求出f(i,j)的值。
class Solution {
public:
int uniquePaths(int m, int n) {
int res[m+1][n+1];
for(int i=1;i<=m;i++)
res[i][1]=1;
for(int i=1;i<=n;i++)
res[1][i]=1;
for(int i=2;i<=m;i++){
for(int j=2;j<=n;j++){
res[i][j]=res[i-1][j]+res[i][j-1];
}
}
return res[m]
;
}
};
(2)单纯从移动来考虑,由于只能向右或者向下移动,所以从(1,1)移动到(m,n)共需要m+n-2步,其中向下m-1步,向右n-1步。所以说,在这m+n-2步中,需要考虑的就是何时向下何时向右,这就变成了排列组合的问题。
class Solution {
public:
int uniquePaths(int m, int n) {
int a=m+n-2,b=max(m,n)-1;
long s=1,t=1;
for(int i=b+1;i<=a;i++)
s*=i;
for(int i=2;i<=a-b;i++)
t*=i;
return s/t;
}
};
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