[LeetCode] 265. Paint House II 粉刷房子
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, costs0 is the cost of painting house 0 with color 0; costs1 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?
解题思路：
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，那么我们用min2对应的值计算，反之我们用min1对应的值，这种解法实际上也包含了求次小值的方法。
State: dp[i][j]
Function: dp[i][j] = costs[i][j] + preMin or costs[i][j] + preSec
Initialize: preMin = 0 , preSec = 0
Return: dp[n][preMin]
Java: Time: O(n), Space: O(1)
public class Solution {
public int minCostII(int[][] costs) {
if(costs != null && costs.length == 0) return 0;
int prevMin = 0, prevSec = 0, prevIdx = -1;
for(int i = 0; i < costs.length; i++){
int currMin = Integer.MAX_VALUE, currSec = Integer.MAX_VALUE, currIdx = -1;
for(int j = 0; j < costs[0].length; j++){
costs[i][j] = costs[i][j] + (prevIdx == j ? prevSec : prevMin);
// 找出最小和次小的，最小的要记录下标，方便下一轮判断
if(costs[i][j] < currMin){
currSec = currMin;
currMin = costs[i][j];
currIdx = j;
} else if (costs[i][j] < currSec){
currSec = costs[i][j];
}
}
prevMin = currMin;
prevSec = currSec;
prevIdx = currIdx;
}
return prevMin;
}
}
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