62. Unique Paths 唯一路径的条数

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.

1.我的直观想法（递归）

当时的直观想法是，当走到最后一格，将总数num加一。

class Solution {
public:
    void uniquePath(int begin, int end, int m, int n, double& num){
        if(begin==m-1 && end == n-1){
            num++;
            return;
        }else if(begin == m-1 || end == n-1){
        if(begin == m-1)
        uniquePath(m-1, end+1, m, n, num);
        if(end == n-1)
        uniquePath(begin+1, n-1, m, n, num);
        }else{
            uniquePath(begin, end+1, m, n, num);
            uniquePath(begin+1, end, m, n, num);
        }
        return;
    }
    int uniquePaths(int m, int n) {
        double num = 0;
        int begin = 0, end = 0;
        if(m == 1 || n == 1)
        return 1;
        uniquePath(begin,end,m,n,num);
        return num;
    }
};

提供几个答案：

3x3->6

3x4->10

5x3->15

2.动态规划

提交后发现会超时，就知道这种方法不对，需要动态规划。那么直接想法是，找一格m*n的格子，每个格子存储着从开始位置到该位置的路径。该位置的路径值就是f[i,j] = f[i-1,j]+f[i,j-1]. 然后初始化时，[0,0~n-1]及[0~m-1,n]都是1，因为只有一条路径到这些位置。

然后再用滚动数组，只用一列，代替m*n的矩阵，就能降低空间复杂度。

class Solution {
public:
    int uniquePaths(int m, int n) {
        double num = 0;
        int begin = 0, end = 0;
        if(m == 1 || n == 1)
        return 1;
        vector<int>vec(n,1);
        while(--m){
        for(int i = 1; i < n; i++){
            vec[i] += vec[i-1]; 
            }
        }
        return vec[n-1];
    }
};