[LintCode]Paint House II

http://www.lintcode.com/en/problem/paint-house-ii/#

刷房子，每个房子刷每种颜色有一个cost[i][j]，相邻两个房子不能刷同一种颜色，求最小cost

要求时间复杂度O(NK)

每次不需要前一状态的所有值，只需要保存前一状态之中最小和第二小的所涂的颜色。

内层循环初始值设置为-1

public class Solution {
/**
* @param costs n x k cost matrix
* @return an integer, the minimum cost to paint all houses
*/
public int minCostII(int[][] costs) {
// Write your code here
if (costs == null || costs.length == 0) {
return 0;
}
int min1 = -1;
int min2 = -1;
int[][] cache = costs.clone();
for (int i = 0; i < costs.length; i++) {
int last1 = min1;
int last2 = min2;
min1 = -1;
min2 = -1;
for (int j = 0; j < costs[0].length; j++) {
if (j != last1) {
cache[i][j] += last1 < 0 ? 0 : cache[i - 1][last1];
} else {
cache[i][j] += last2 < 0 ? 0 : cache[i - 1][last2];
}
if (min1 < 0 || cache[i][min1] > cache[i][j]) {
min2 = min1;
min1 = j;
} else if (min2 < 0 || cache[i][min2] > cache[i][j]) {
min2 = j;
}
}
}
return cache[costs.length - 1][min1];
}
}