[LeetCode]Paint Fence

Paint Fence

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.

分析

典型的DP题。因为这里是只允许两个连续的fence时同样的颜色，所以对于dp[i]而言，其状态不仅与dp[i-1]有关，也与dp[i-2]有关。

我们可以分两种情况考虑，一种是当前fence颜色与上个fence颜色不同，一种是当前fence颜色与上个fence颜色相同。对于前者，显然dp[i] = (k - 1) * dp[i - 1]。 对于后者，由于题目要求不能连续三个fence颜色相同，我们只要保证上上个fence与当前fence颜色不同即可，此时dp[i] = 1 * (k - 1) * dp[i - 2]。

所以 dp[i] = (k - 1) * dp[i - 1] + (k - 1) * dp[i - 2]

当然, 用两个变量代替int数组，时间复杂度是可以降至O(1)的。

另外，这道题也可以有Follow up比如题目改成最多允许3个fence有相同颜色，结果怎么样？

用上面类似的分析，可得

dp[i] = dp[i - 1] * (k - 1) + dp[i - 2] * (k - 1) * 1 + dp[i - 3] * (k - 1) * 1 * 1 

复杂度

time: O(n), space: O(n)

代码

public class Solution {
    public int numWays(int n, int k) {
        if (n == 0) return 0;
        if (n == 1) return k;
        int[] dp = new int[n];
        dp[0] = k;
        dp[1] = k * (k - 1) + k;
        for (int i = 2; i < n; i++) {
            dp[i] = dp[i - 1] * (k - 1) + dp[i - 2] * (k - 1);
        }
        return dp[n - 1];
    }
}