refactor 265

Author: fishercoder1534

src/main/java/com/fishercoder/solutions/_265.java

@@ -1,65 +1,45 @@
 package com.fishercoder.solutions;
 
-/**
- * 265. Paint House II
- *
- * 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?
- */
 public class _265 {
 
-	public static class Solution1 {
-		public int minCostII(int[][] costs) {
-			if (costs == null || costs.length == 0) {
-				return 0;
-			}
-
-			int n = costs.length;
-			int 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;
-			int min2 = -1;
-
-			for (int i = 0; i < n; i++) {
-				int last1 = min1;
-				int 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];
-		}
-	}
+    public static class Solution1 {
+        public int minCostII(int[][] costs) {
+            if (costs == null || costs.length == 0) {
+                return 0;
+            }
+
+            int n = costs.length;
+            int 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;
+            int min2 = -1;
+
+            for (int i = 0; i < n; i++) {
+                int last1 = min1;
+                int 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];
+        }
+    }
 
 }