[LeetCode] 276. Paint Fence_Easy tag: Dynamic Programming

There is a fence with n posts, each post can be painted with one of the k colors.

You have to paint all the posts such that no more than two adjacent fence posts have the same color.

Return the total number of ways you can paint the fence.

Note:
n and k are non-negative integers.

Example:

Input: n = 3, k = 2
Output: 6
Explanation: Take c1 as color 1, c2 as color 2. All possible ways are:

            post1  post2  post3      
 -----      -----  -----  -----       
   1         c1     c1     c2 
   2         c1     c2     c1 
   3         c1     c2     c2 
   4         c2     c1     c1  
   5         c2     c1     c2
   6         c2     c2     c1

思路是利用两个dp, 一个same, 一个dif, same[i] 表明i 跟i-1 两个fence 的颜色是一样的, 而dif[i] 表明i 跟 i-1 两个 fence的颜色是不一样的.

init:   same[0] = same[1] = k

        dif[0] = k, dif[1] = k*(k-1)

same[i] = dif[i-1]

dif[i] = (same[i-1] + dif[i-1]) *(k-1)

Code    T: O(n)   S; O(1)  using rolling array

class Solution:
    def numWays(self, n, k):
        if n == 0: return 0
        if n == 1: return k
        same, dif = [0]*2, [0]*2
        same[0] = same[1] = k
        dif[0], dif[1] = k, k*(k-1)
        for i in range(2, n):
            same[i%2] = dif[i%2-1]
            dif[i%2] = (k-1)*(same[i%2-1] + dif[i%2-1])
        return same[(n-1)%2] + dif[(n-1)%2]

[LeetCode] 276. Paint Fence_Easy tag: Dynamic Programming

原文：https://www.cnblogs.com/Johnsonxiong/p/9545923.html