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;
}
}