[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> f(m,vector(n,0));

        for(int i=0;i

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 f(n,0);
    f[0]=1;

    for(int i=0;i