Leetcode(62)Unique Paths
来源:互联网 发布:巨星知我心 by凌豹姿 编辑:程序博客网 时间:2024/06/05 21:53
题目:从左上角到右下角,只能往右走或者往下走,有多少种不同的走法。
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?
分析:动态规划,设f[i][j]表示从点(i,j)到终点有多少种不同的走法。那么f[i][j]=f[i+1][j]+f[i][j+1].
边界条件:f[m-1][n-1]=1
方向:从向往上,从右往左
复杂度:O(n²)。
用动态规划很慢呀,其实这道题可以直接计算。因为每种走法有m-1次向右,n-1次向下,所以其实就是从m+n-2次移动中选择m-1次为向右移,其他位向左移。
所以答案是C(m+n-2,m-1)。
代码(动态规划):
class Solution {public: int uniquePaths(int m, int n) { if(m==0||n==0) return 0; if(m==1||n==1) return 1; vector<vector<int> > f(m,vector<int>(n, 0)); f[m-1][n-1]=1; for(int j=n-1;j>=0;j--) for(int i=m-1;i>=0;i--) { if(i+1<m) f[i][j]+=f[i+1][j]; if(j+1<n) f[i][j]+=f[i][j+1]; } return f[0][0]; }};排列组合:
class Solution {public: int uniquePaths(int m, int n) { double res=1; int N=m+n-2; int k=1; while(k<=n-1) { res=res*(N-(n-1)+k)/k; k++; } return (int)res; }};
0 0
- LeetCode(62)Unique Paths
- Leetcode(62)Unique Paths
- [leetcode 62] Unique Paths
- LeetCode(62): Unique Paths
- [LeetCode 62]Unique Paths
- leetcode || 62、Unique Paths
- Unique Paths - LeetCode 62
- leetcode-62 Unique Paths
- Leetcode[62]-Unique Paths
- leetcode 62:Unique Paths
- Leetcode #62 Unique Paths
- leetcode 62: Unique Paths
- Leetcode#62||Unique Paths
- LeetCode 62: Unique Paths
- 【leetcode】【62】Unique Paths
- leetcode #62 Unique Paths
- leetcode 62:Unique Paths
- 【LEETCODE】62-Unique Paths
- 让你的代码量减少3倍!使用kotlin开发Android(二) --秘笈!扩展函数
- 48. 减少 DNS 查找(9)
- 文章标题
- 分页技术
- 操作系统
- Leetcode(62)Unique Paths
- 关于Application类不得不说的事情
- 文章标题
- HDU 5904 BestCoder Round #88 Find Q (统计Q!)
- 折线分割平面
- 2.长川科技
- 11gRAC搭建单机Dataguard
- poj1003
- @media:device-width、orientation