problem:
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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题意:给定一个 mxn的矩阵,求从(0,0)到(m-1,n-1)的路径数,每次只能往x 、y增长的方向移动
thinking:
(1)这道题相当经典,为什么这样说,且听分析
看到题第一反应想到用DFS遍历每一条路径,计个数即可。但是DFS的时间复杂度是路径数total 的线性表达式,提交显示 TLE
(2)抖机灵想到用递归,f(m,n)=f(m-1,n)+f(m,n-1),当m或n为1时,路径数为1。 但是,递归法的时间复杂度同DFS法一样,提交也是TLE
(3)正确的解法是DP,状态转移方程 同递归法一样:a[m][n] = a[m-1][n]+a[m][n-1],采用自底向上的思路,时间复杂度和空间复杂度都为O(m*n)
code:
DFS法: TLE
class Solution { public: int uniquePaths(int m, int n) { int count=0; int x=0,y=0; dfs(x,y,m,n,count); return count; } protected: void dfs(int x, int y, int m, int n, int &count) { if(x==m-1 && y==n-1) { count++; return; } if(x<m-1 && y< n-1) { dfs(x+1,y,m,n,count); dfs(x,y+1,m,n,count); } else if(x==m-1 && y<n-1) dfs(x,y+1,m,n,count); else dfs(x+1,y,m,n,count); } };
递归法:TLE
class Solution { public: int uniquePaths(int m, int n) { return f(m,n); } protected: int f(int m,int n) { if(m==1 || n==1) return 1; else return f(m-1,n)+f(m,n-1); } };
DP法:accepted
class Solution { public: int uniquePaths(int m, int n) { vector<vector<int> > a(m, vector<int>(n)); for(int i = 0; i < n; i++) a[0][i] = 1; for(int i = 0; i < m; i++) a[i][0] = 1; for(int i = 1; i < m; i++) for(int j = 1; j < n; j++) a[i][j] = a[i-1][j] + a[i][j-1]; return a[m-1][n-1]; } };