Lintcode: Minimum Adjustment Cost

Given an integer array, adjust each integers so that the difference of every adjcent integers are not greater than a given number target.

If the array before adjustment is A, the array after adjustment is B, you should minimize the sum of |A[i]-B[i]| 

Note
You can assume each number in the array is a positive integer and not greater than 100

Example
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.

这道题要看出是背包问题，不容易，跟FB一面 paint house很像，比那个难一点

定义res[i][j] 表示前 i个number with 最后一个number是j，这样的minimum adjusting cost

如果第i-1个数是j, 那么第i-2个数只能在[lowerRange, UpperRange]之间，lowerRange=Math.max(0, j-target), upperRange=Math.min(99, j+target),

这样的话，transfer function可以写成：

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)));

}

public class Solution {
    /**
     * @param A: An integer array.
     * @param target: An integer.
     */
    public int MinAdjustmentCost(ArrayList<Integer> A, int target) {
        // write your code here
        int[][] res = new int[A.size()+1][100];
        for (int j=0; j<=99; j++) {
            res[0][j] = 0;
        }
        for (int i=1; i<=A.size(); i++) {
            for (int j=0; j<=99; j++) {
                res[i][j] = Integer.MAX_VALUE;
                int lowerRange = Math.max(0, j-target);
                int upperRange = Math.min(99, j+target);
                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)));
                }
            }
        }
        int result = Integer.MAX_VALUE;
        for (int j=0; j<=99; j++) {
            result = Math.min(result, res[A.size()][j]);
        }
        return result;
    }
}