265. Paint House II

  /*
   * 265. Paint House II
   * 2016-6-24 by Mingyang
   * 这个题目说简单也简单，说难也难，因为简单在于他可以用n^3很容易做出来，这样的话就不用担心
   * 但是要用平方做出来非常复杂
   * 第一种跟I一样：
   * Define dp[n][k], where dp[i][j] means for house i with color j the minimum cost. 
   * Initial value: dp[0][j] = costs[0][j]. For others, dp[i][j] = Integer.MAX_VALUE;, i >= 1
   * Transit function: dp[i][j] = Math.min(dp[i][j], dp[i - 1][k] + cost[i][j]), where k != j.
   * Final state: Min(dp[n - 1][k]).
   */
      public int minCostII(int[][] costs) {
            if (costs == null || costs.length == 0) {
                return 0;
            }         
            int n = costs.length;
            int k = costs[0].length;             
            // dp[i][j] means the min cost painting for house i, with color j
            int[][] dp = new int[n][k];         
            // Initialization
            for (int i = 0; i < k; i++) {
                dp[0][i] = costs[0][i];
            }             
            for (int i = 1; i < n; i++) {
                for (int j = 0; j < k; j++) {
                    dp[i][j] = Integer.MAX_VALUE;
                    for (int m = 0; m < k; m++) {
                        if (m != j) {
                            dp[i][j] = Math.min(dp[i - 1][m] + costs[i][j], dp[i][j]);
                        }
                    }
                }
            }             
            // Final state
            int minCost = Integer.MAX_VALUE;
            for (int i = 0; i < k; i++) {
                minCost = Math.min(minCost, dp[n - 1][i]);
            }         
            return minCost;
        }
      //这就是高手的平方做法，就是用两个index分别来存这一行里面，最小的和第二小的颜色的值，然后不断地更新
      public int minCostII2(int[][] costs) {
            if (costs == null || costs.length == 0) return 0;
            int n = costs.length, k = costs[0].length;
            // min1 is the index of the 1st-smallest cost till previous house
            // min2 is the index of the 2nd-smallest cost till previous house
            int min1 = -1, min2 = -1;
            for (int i = 0; i < n; i++) {
                int last1 = min1, last2 = min2;
                min1 = -1; min2 = -1;
                for (int j = 0; j < k; j++) {
                    if (j != last1) {
                        // current color j is different to last min1
                        costs[i][j] += last1 < 0 ? 0 : costs[i - 1][last1];
                    } else {
                        costs[i][j] += last2 < 0 ? 0 : costs[i - 1][last2];
                    }
         // find the indices of 1st and 2nd smallest cost of painting current house i
                    if (min1 < 0 || costs[i][j] < costs[i][min1]) {
                        min2 = min1; min1 = j;
                    } else if (min2 < 0 || costs[i][j] < costs[i][min2]) {
                        min2 = j;
                    }
                }
            }
            return costs[n - 1][min1];
        }