LeetCode 276. Paint Fence

最新推荐文章于 2024-04-23 10:23:23 发布

原创最新推荐文章于 2024-04-23 10:23:23 发布 · 356 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 题解专栏收录该内容

288 篇文章

订阅专栏

本文详细解析了LeetCode中的第276题——涂色栅栏。题目要求给一个长度为n的栅栏涂色，每种颜色至少使用一次，求所有可能的涂色方案数。通过动态规划的方法，我们可以解决这个问题。文章深入浅出地介绍了动态规划的思路和实现代码，适合想要提升算法能力的开发者阅读。

#include <iostream>
#include <vector>
using namespace std;

/*
  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 colors.
  Return the total number of ways you can paint the fence. 
  Note: n and k are non-negative integers.
*/

/*
  It is better to use an example to illustrate the question.
  if n == 0, return 0 method.
  if n == 1, return k method (1 post can have k colors)
  
  // k-1 is the colors for second post different from the first one.
  // 1 is the color that  same as first post.
  if n == 2, return k * (k - 1) + k * 1. 

  // (k * 1 * (k-1) + k * (k - 1) * (k - 1)) is the choice that different from second post.
  // (k * (k-1) + k * 1) * 1, is the choice that same color as second post.
  if n == 3, return (k * 1 * (k-1) + k * (k - 1) * (k - 1)) + (k * (k-1) + k * 1) * 1 
*/
int numOfWays(int n, int k) {
  if(n == 0) return 0;
  if(n == 1) return k;
  int same_color = k;
  int different_color = k * (k - 1);
  for(int i = 3; i <= n; ++i) {
    int tmp = different_color;
    different_color = different_color * (k-1) + same_color * (k-1);
    same_color = tmp * 1;
  }
  return different_color + same_color;
}

int main(void) {
  cout << numOfWays(3, 2) << endl;
}