1 Given an integer array, adjust each integers so that the difference of every adjcent integers are not greater than a given number target.2 3 If the array before adjustment is A, the array after adjustment is B, you should minimize the sum of |A[i]-B[i]| 4 5 Note6 You can assume each number in the array is a positive integer and not greater than 1007 8 Example9 Given [1,4,2,3] and target=1, one of the solutions is [2,3,2,3], the adjustment cost is 2 and it's minimal. Return 2.

這道題要看出是背包問題，不容易，跟FB一面 paint house很像，比那個難一點

定義res[i][j] 表示前 i個number with 最后一個number是j，這樣的minimum adjusting cost

如果第i-1個數(shù)是j, 那么第i-2個數(shù)只能在[lowerRange, UpperRange]之間，lowerRange=Math.max(0, j-target), upperRange=Math.min(99, j+target),

這樣的話，transfer function可以寫成：

for (int p=lowerRange; p<= upperRange; p++) {

　　res[i][j] = Math.min(res[i][j], res[i-1][p] + Math.abs(j-A.get(i-1)));

}

 1 public class Solution { 2     /** 3      * @param A: An integer array. 4      * @param target: An integer. 5      */ 6     public int MinAdjustmentCost(ArrayList<Integer> A, int target) { 7         // write your code here 8         int[][] res = new int[A.size()+1][100]; 9         for (int j=0; j<=99; j++) {10             res[0][j] = 0;11         }12         for (int i=1; i<=A.size(); i++) {13             for (int j=0; j<=99; j++) {14                 res[i][j] = Integer.MAX_VALUE;15                 int lowerRange = Math.max(0, j-target);16                 int upperRange = Math.min(99, j+target);17                 for (int p=lowerRange; p<=upperRange; p++) {18                     res[i][j] = Math.min(res[i][j], res[i-1][p]+Math.abs(j-A.get(i-1)));19                 }20             }21         }22         int result = Integer.MAX_VALUE;23         for (int j=0; j<=99; j++) {24             result = Math.min(result, res[A.size()][j]);25         }26         return result;27     }28 }