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.
Example:
Input: [[1,5,3],[2,9,4]]
Output: 5
Explanation: Paint house 0 into color 0, paint house 1 into color 2. Minimum cost: 1 + 4 = 5;
Or paint house 0 into color 2, paint house 1 into color 0. Minimum cost: 3 + 2 = 5.
Follow up:
Could you solve it in O(nk) runtime?
思路:每次都只需要保存前一个房子涂不同颜色的最小值即可。时间复杂度为nkk
public int minCostII(int[][] costs) {
if (costs.length == 0 || costs[0].length == 0) return 0;
int house = costs.length;
int color = costs[0].length;
int[] dp = new int[color];
dp = costs[0];
for (int i = 1; i < house; i++) {
int[] cur = new int[color];
for (int j = 0; j < color; j++) {
int min = Integer.MAX_VALUE;
for (int c = 0; c < color; c++) {
if (c != j) {
min = Math.min(min, costs[i][j] + dp[c]);
}
}
cur[j] = min;
}
dp = cur;
}
int res = Integer.MAX_VALUE;
for (int i = 0; i < dp.length; i++) {
res = Math.min(res, dp[i]);
}
return res;
}
简化版:时间复杂度为n*k,每次只需保存前一个房子的最小代价和涂色以及第二小代价
public int minCostII(int[][] costs) {
if(costs == null || costs.length == 0 || costs[0].length == 0) return 0;
int n = costs.length, k = costs[0].length;
if(k == 1) return (n==1? costs[0][0] : -1);
int prevMin = 0, prevMinInd = -1, prevSecMin = 0;//prevSecMin always >= prevMin
for(int i = 0; i<n; i++) {
int min = Integer.MAX_VALUE, minInd = -1, secMin = Integer.MAX_VALUE;
for(int j = 0; j<k; j++) {
int val = costs[i][j] + (j == prevMinInd? prevSecMin : prevMin);
if(minInd< 0) {min = val; minInd = j;}//when min isn't initialized
else if(val < min) {//when val < min,
secMin = min;
min = val;
minInd = j;
} else if(val < secMin) { //when min<=val< secMin
secMin = val;
}
}
prevMin = min;
prevMinInd = minInd;
prevSecMin = secMin;
}
return prevMin;
}