Leetcode: Unique Paths

标签：leetcode   动态优化   

题目：

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.

思路分析：

又是动态规划问题。

开一个f[m][n]的数组，数组元素初始化为1，递推公式f[i][j] = f[i-1][j] + f[i][j-1]，空间时间复杂度O(m*n)。

（可以将f[m][n]理解成为从f[0][0]到达f[m][n]的路径个数。那很自然的就会f[i][j] = f[i-1][j] + f[i][j-1]。有感觉递推公式还不是能很好想出来的，继续加强训练吧！）

C++参考代码：

class Solution
{
public:
    int uniquePaths(int m, int n)
    {
        //将vector中的元素初始化为1
        vector<vector<int>> v(m, vector<int>(n, 1));
        for (int i = 1; i < m; ++i)
        {
            for (int j = 1; j < n; ++j)
            {
                v[i][j] = v[i - 1][j] + v[i][j - 1];
            }
        }
        return v[m - 1][n - 1];
    }
};

所以利用杨辉三角，开一个f[n]的数组，数组元素初始化为1，递推公式f[i]+=f[i-1]，空间时间复杂度O(n)。

class Solution
{
public:
    int uniquePaths(int m, int n)
    {
        vector<int> v(n, 1);
        for (int i = 1; i < m; ++i)
        {
            for (int j = 1; j < n; ++j)
            {
                v[j] += v[j - 1];
            }
        }
        return v[n - 1];
    }
};

Leetcode: Unique Paths

标签：leetcode   动态优化   

原文地址：http://blog.csdn.net/theonegis/article/details/44968639