[LeetCode] Unique Paths

Unique Paths

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Question Solution 

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.

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 Array Dynamic Programming

 

思路一： 动态规划， f[i][j] = f[i-1][j] + f[i][j-1], 且f[i][0] = 0, f[0][j]=0, f[1][1] =1 ,时间复杂度O(m*n)，空间复杂度O(m*n)

class Solution {
    public:
        int uniquePaths(int m, int n)
        {   
            vector<int> row(n + 1, 0); 
            vector<vector<int> > f(m + 1, row);

            f[1][1] = 1;

            for(int i = 1; i <= m; i ++) 
            {   
                for(int j = 1; j <= n; j ++) 
                {   
                    if(i == 1 && j == 1)
                        continue;
                    f[i][j] = f[i-1][j] + f[i][j-1];
                }   
            }   

            return f[m][n];
        }   
};

 思路二：空间复杂度还可以优化，将f[][] 改成 f[],优化后，时间复杂度不变，空间复杂度O(n)

 

class Solution {
    public:
#if 0
        int uniquePaths(int m, int n)
        {
            vector<int> row(n + 1, 0);
            vector<vector<int> > f(m + 1, row);

            f[1][1] = 1;

            for(int i = 1; i <= m; i ++)
            {
                for(int j = 1; j <= n; j ++)
                {
                    if(i == 1 && j == 1)
                        continue;
                    f[i][j] = f[i-1][j] + f[i][j-1];
                }
            }

            return f[m][n];
        }
#endif
        int uniquePaths(int m, int n)
        {
            vector<int> f(n + 1, 0);

            f[1] = 1;

            for(int i = 1; i <= m; i ++)
            {
                for(int j = 1; j <= n; j ++)
                {
                    if(i == 1 && j == 1)
                        continue;
                    f[j] = f[j] + f[j-1];
                }
            }

            return f[n];
        }
};