62. Unique Paths
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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.
这一题的突破点在于数学上。
如果用递归的方法,每走一步,都有两个选择:往左or往右。那么时间复杂度为2^(min(n,m)-1),是指数类型,我们放弃这种做法。
如果我们仔细思考,会发现,在一张mXn图中,我们必须往右走n-1步,往下走m-1步。那么,我们可以通过排列组合知道
结果 = (m+n-2)! / (m-1)!(n-1)!
但是,如果m,n过大,直接算的话会发生溢出。例如当m=20,n=20,则(m+n-2)! = 38!,必定溢出。
那么我们可以通过动态规划来进行。
令f(m,n)为在mXn地图需要走的步数。显然 f(m, n) = (m+n-2)! / (m-1)!(n-1)!
则 f(m, n-1) = (m+n-3) / (m-1)!(n-2)!
得到 f(m, n) = f(m, n-1)*(m+n-2) / (n-1)
通过这个递推式我们就可以写出我们的代码:
int uniquePaths(int m, int n) {
if (m == 1 || n == 1) return 1; //当n = 1或 m = 1,只有一种路径
int a = 1; //a = f(m, 1)
for (int i = 2; i <= n; i++)
a = a*(m+i-2)/(i-1); //f(m, n) = f(m, n-1)*(m+n-2) / (n-1)
return a;
}
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