Question
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 each house with a certain color is represented by a n x k cost matrix. For example, costs[0][0] is the cost of painting house 0 with color 0; costs[1][2] is the cost of painting house 1 with color 2, and so on... Find the minimum cost to paint all houses.
Note:
All costs are positive integers.
Follow up:
Could you solve it in O(nk) runtime?
Solution
这里时间复杂度是O(nk)，说明要求我们用O(k)的时间计算每一层的新的cost。
cost‘[i] = costs[m][i] + min{cost[0], cost[1], ..., cost[i - 1], cost[i + 1], ..., cost[k - 1]}
原想法是对每一个i重新计算，时间复杂度是O(k2)。包含了大量的重复计算
其实我们只需求出cost[]序列的最小值和第二小的值。Time complexity O(k)
1 public class Solution {
2 public int minCostII(int[][] costs) {
3 if (costs == null || costs.length < 1) {
4 return 0;
5 }
6 int m = costs.length, k = costs[0].length;
7 int[] cost = new int[k];
8 int[] tmp = new int[k];
9 for (int i = 0; i < k; i++) {
10 cost[i] = costs[m - 1][i];
11 }
12 for (int i = m - 2; i >= 0; i--) {
13 // calculate most and second minimum number
14 int[] min = calcMin(cost);
15 for (int j = 0; j < k; j++) {
16 // if cost[j] is minimum, then add second minimum with costs[i][j]
17 if (cost[j] == min[0]) {
18 cost[j] = min[1] + costs[i][j];
19 } else {
20 // if cost[j] is not minimum, then add minimum with costs[i][j]
21 cost[j] = min[0] + costs[i][j];
22 }
23 }
24 }
25 if (k < 2) {
26 return cost[0];
27 }
28 int[] result = calcMin(cost);
29 return result[0];
30 }
31
32 private int[] calcMin(int[] nums) {
33 if (nums == null || nums.length < 2) {
34 return new int[0];
35 }
36 int[] mins = new int[2];
37 mins[0] = Math.min(nums[0], nums[1]);
38 mins[1] = Math.max(nums[0], nums[1]);
39 for (int i = 2; i < nums.length; i++) {
40 if (nums[i] < mins[0]) {
41 mins[1] = mins[0];
42 mins[0] = nums[i];
43 } else if (nums[i] < mins[1]) {
44 mins[1] = nums[i];
45 }
46 }
47 return mins;
48 }
49 }