LC256. Paint House
There are a row of n houses, each house can be painted with one of the three colors: red, blue or 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 color.
The cost of painting each house with a certain color is represented by a n x 3 cost matrix. For example, costs[0][0] is the cost of painting house 0 with color red; costs[1][2] is the cost of painting house 1 with color green, and so on… Find the minimum cost to paint all houses.
Note: All costs are positive integers.
DP解，每一个状态包括三种颜色，最后取最小值。
public class Solution {
public int minCost(int[][] costs) {
if(costs == null || costs.length == 0) return 0;
int[] cache = new int[3];
int[] dp = new int[3];
for(int i = 0; i < 3; i++) cache[i] = costs[0][i];
for(int i = 1; i < costs.length; i++){
dp[0] = Math.min(cache[1], cache[2]) + costs[i][0];
dp[1] = Math.min(cache[2], cache[0]) + costs[i][1];
dp[2] = Math.min(cache[1], cache[0]) + costs[i][2];
cache = dp.clone();
}
return Math.min(Math.min(cache[0], cache[1]), cache[2]);
}
}
O(nk), O(k)
LC265. 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?
唔，那我把dp数组设大一点好了。
public class Solution {
public int minCostII(int[][] costs) {
if(costs == null || costs.length == 0) return 0;
int n = costs.length, k = costs[0].length;
int[] dp = new int[k];
int[] cache = new int[k];
for(int i = 0; i < k; i++){
cache[i] = costs[0][i];
}
for(int i = 1; i < n; i++){
for(int j = 0; j < k; j++) dp[j] = findMin(cache, j) + costs[i][j];
cache = dp.clone();
}
return findMin(cache, -1);
}
private int findMin(int[] array, int pos){
int res = Integer.MAX_VALUE;
for(int i = 0; i < array.length; i++){
if(i != pos) res = Math.min(res, array[i]);
}
return res;
}
}
这是一个naive的O(nk^2), O(k)解法
Follow up
还是看了discuss，O(nk)+O(1)解法，每个房子只要找到最小的两个值就可以了，除非最小值和前一个颜色相同用次小值，否则用最小值。PS.这一优化方法在背包九讲中有介绍过。
public class Solution {
public int minCostII(int[][] costs) {
if(costs == null || costs.length == 0) return 0;
int n = costs.length, k = costs[0].length;
int min1 = 0, min2 = 0, index = -1;
for(int i = 0; i < n; i++){
int m1 = Integer.MAX_VALUE, m2 = Integer.MAX_VALUE, idx1 = -1;
for (int j = 0; j < k; j++) {
int cost = costs[i][j] + (j != index? min1 : min2);
if (cost < m1) { // cost < m1 < m2
m2 = m1; m1 = cost; idx1 = j;
} else if (cost < m2) { // m1 < cost < m2
m2 = cost;
}
} // for j
min1 = m1;
min2 = m2;
index = idx1;
} // for i
return min1;
}
}