[LeetCode]-Unique Paths 矩阵中求两点间所有路线条数

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?
方法1：直接采用递归的方式。小集合可以通过，不能通过大集合。

class Solution {
public:
int uniquePaths(int m, int n) {
int paths=0;
compute(m,n,0,0,paths);
return paths;
}

void compute(int m,int n,int i,int j,int &paths){
if(i==m-1 && j==n-1){
paths++;
return;
}
if(i>=m || j>=n)return;
compute(m,n,i+1,j,paths);
compute(m,n,i,j+1,paths);
}
};

方法二：采用动态规划的方式。自底向上的解决问题。

设状态f[i][j] 表示(0,0)到( i , j )的路线条数。状态转移方程：f[i][j]=f[i-1][j]+ f[i][j-1]

class Solution {
public:
int uniquePaths(int m, int n) {
vector<vector<int>> f(m,vector<int>(n,0));

for(int i=0;i<m;i++)
f[i][0]=1;
for(int i=0;i<n;i++)
f[0][i]=1;

for(int i=1;i<m;i++)
for(int j=1;j<n;j++)
f[i][j]=f[i-1][j]+f[i][j-1];
return f[m-1][n-1];
}
};

但是花费了较大开销，可以使用滚动数组的方式节省空间开销。

方法三：动态规划+滚动数组

class Solution {
public:
int uniquePaths(int m, int n) {

vector<int> f(n,0);
f[0]=1;

for(int i=0;i<m;i++)
for(int j=1;j<n;j++)
//左边的f[j]，表示是本轮i，则为f[i][j]
//f[j],表示是上一轮i-1,则代表f[i-1][j]
//f[j-1],表示是本轮i,则代表f[j-1][j]
f[j]=f[j-1]+f[j];
return f[n-1];
}
};