[LeetCode] 265. Paint House II 粉刷房子之二 There are a row ofnhouses, each house can be painted with one of thekcolors. The cost of painting each house with a certain color is different. You have to paint all the houses such that no two adjacent houses have the same color. The co...
Leetcode: Paint House II There are a row of n houses, each house can be painted with one of the k colors. The cost of painting each house with a certain color is different. You have to paint all the houses such that no two adjacent houses have the same color. The cost of painting...
ans != MAX: ans 即为满足题意的所有染色方案中,花费最小的费用 DP 常见的三种优化方式见LeetCode 583 这题的思路。 本题可以采用滚动数组的方式进行优化,因为每一行的状态依赖上一行的所有状态,所以无法采用其他两种方式进行优化。 使用滚动数组的话,能将空间复杂度从O(m * target * n) 优化为 O(target *...
https://leetcode.com/problems/paint-fence/ 不能有两个以上相邻的fence是相同color。 第三根柱子要么根第一个柱子不是一个颜色,要么跟第二根柱子不是一个颜色 classSolution{public:intnumWays(intn,intk){if(n==0||k==0)return0;vector<int>dp(n);dp[0]=k;dp[1]=k*k;for(inti=2;i<n;i++...
Linkedin Interview - Paint House with Colors There are a row of houses, each house can be painted with three colors red, blue and green. The cost of painting each house with a certain color is different. You have to paint all the houses such that no two adjacent houses have the same ...
Return the total number of ways you can paint the fence. [暴力解法]: 时间分析: 空间分析: [奇葩输出条件]: [奇葩corner case]: 可以2个柱子颜色相同,则起点从2开始, 之前的所有情况,包括0 和 1 都要分别列出来写 [思维问题]: 列了几步以后觉得列不完,没思路:从小到大,反正只有相同、不相同两种情...
Or paint house 0 into color 2, paint house 1 into color 0. Minimum cost: 3 + 2 = 5. Follow up: Could you solve it in O(nk) runtime? 答案 直接改编Paint House题的答案 O(nk^2) classSolution{publicintminCostII(int[][]costs){if(costs.length==0)return0;intn=costs.length,k=cost...
There are a row ofnhouses, each house can be painted with one of thekcolors. The cost of painting each house with a certain color is different. You have to paint all the houses such that no two adjacent houses have the same color. ...
这道题是Paint House的拓展,这题的解法的思路还是用DP,那道题只让用红绿蓝三种颜色来粉刷房子,而这道题让我们用k种颜色,这道题不能用之前那题的Math.min方法了,会TLE。只要把最小和次小的都记录下来就行了,用preMin和PreSec来记录之前房子的最小和第二小的花费的颜色,如果当前房子颜色和min1相同,那么我们...
256. 265. Paint House https://leetcode.com/problems/paint-house/description/ image.png 因为有3种颜色,所以在建立DP的时候,需要维护3个状态,就是最后一个以什么颜色的HOSUE,最少要花多少。 dp[i][0] = min(dp[i-1 ][1],dp[i-1 ][2])+cost[i][0]转移方程就出来了。