Paint House II 解答

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)

public class Solution {
public int minCostII(int[][] costs) {
if (costs == null || costs.length < 1) {
return 0;
}
int m = costs.length, k = costs[0].length;
int[] cost = new int[k];
int[] tmp = new int[k];
for (int i = 0; i < k; i++) {
cost[i] = costs[m - 1][i];
}
for (int i = m - 2; i >= 0; i--) {
// calculate most and second minimum number
int[] min = calcMin(cost);
for (int j = 0; j < k; j++) {
// if cost[j] is minimum, then add second minimum with costs[i][j]
if (cost[j] == min[0]) {
cost[j] = min[1] + costs[i][j];
} else {
// if cost[j] is not minimum, then add minimum with costs[i][j]
cost[j] = min[0] + costs[i][j];
}
}
}
if (k < 2) {
return cost[0];
}
int[] result = calcMin(cost);
return result[0];
}

private int[] calcMin(int[] nums) {
if (nums == null || nums.length < 2) {
return new int[0];
}
int[] mins = new int[2];
mins[0] = Math.min(nums[0], nums[1]);
mins[1] = Math.max(nums[0], nums[1]);
for (int i = 2; i < nums.length; i++) {
if (nums[i] < mins[0]) {
mins[1] = mins[0];
mins[0] = nums[i];
} else if (nums[i] < mins[1]) {
mins[1] = nums[i];
}
}
return mins;
}
}