[leetcode] Unique Paths

最新推荐文章于 2024-07-25 09:04:50 发布

Cabinathor

最新推荐文章于 2024-07-25 09:04:50 发布

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分类专栏： Leetcode

本文链接：https://blog.youkuaiyun.com/lmxmimihuhu/article/details/46626193

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本文深入探讨了机器人从网格的左上角到达右下角的不同路径数量问题，通过多种算法实现并优化解决方案，最终利用组合数学原理进行高效计算。

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From : https://leetcode.com/problems/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?

Note: m and n will be at most 100.

Solution 1:

grid[m][n]的数组，grid[i][j] =grid[i-1][j] + grid[i][j-1]，空间时间复杂度O(m*n)。用滚动数组空间复杂度可降为O(n).

class Solution {
public:
    int uniquePaths(int m, int n) {
        vector<vector<int>> grid(m, vector<int>(n));
        for(int i=0; i<m; i++) grid[i][0]=1;
        for(int j=0; j<n; j++) grid[0][j]=1;
        for(int i=1; i<m; i++) {
            for(int j=1; j<n; j++) {
                grid[i][j] = grid[i-1][j]+grid[i][j-1];
            }
        }
        return grid[m-1][n-1];
    }
};

优化得到：

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

Solution 2:

向下走m-1，向右走n-1。即从m+n-2步中，找出m步往下，其余步向右即可。组合数 C(m+n-2, m-1); 组合数可以用杨辉三角求。

class Solution {
public:
    int uniquePaths(int m, int n) {
        vector<int> line;
        if(m==1 || n==1) return 1;
        n = m+n-2;
        line.push_back(1);
        for(int i=0; i<n; i++) {
            line.push_back(1);
            for(int j=line.size()-2; j>0; j--) {
                line[j] = line[j]+line[j-1];
            }
        }
        return line[m-1];
    }
};

最后我笑了。

class Solution {
public:
    int uniquePaths(int m, int n) {
        double dom = 1;
        double dedom = 1;
        int small = m<n? m-1:n-1;
        int big = m<n? n-1:m-1;
        for(int i=1;i<=small;i++) {
            dedom *= i;
            dom *= small+big+1-i;
        }
        return (int)(dom/dedom);
    }
};