forked from fishercoder1534/Leetcode
-
Notifications
You must be signed in to change notification settings - Fork 0
/
_265.java
65 lines (56 loc) · 1.71 KB
/
_265.java
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
package com.fishercoder.solutions;
/**
* 265. 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 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?
*/
public class _265 {
public static class Solution1 {
public int minCostII(int[][] costs) {
if (costs == null || costs.length == 0) {
return 0;
}
int n = costs.length;
int k = costs[0].length;
// min1 is the index of the 1st-smallest cost till previous house
// min2 is the index of the 2nd-smallest cost till previous house
int min1 = -1;
int min2 = -1;
for (int i = 0; i < n; i++) {
int last1 = min1;
int last2 = min2;
min1 = -1;
min2 = -1;
for (int j = 0; j < k; j++) {
if (j != last1) {
// current color j is different to last min1
costs[i][j] += last1 < 0 ? 0 : costs[i - 1][last1];
} else {
costs[i][j] += last2 < 0 ? 0 : costs[i - 1][last2];
}
// find the indices of 1st and 2nd smallest cost of painting current house i
if (min1 < 0 || costs[i][j] < costs[i][min1]) {
min2 = min1;
min1 = j;
} else if (min2 < 0 || costs[i][j] < costs[i][min2]) {
min2 = j;
}
}
}
return costs[n - 1][min1];
}
}
}