Lintcode: Minimum Adjustment Cost
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 Note 6 You can assume each number in the array is a positive integer and not greater than 100 7 8 Example 9 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)));
}
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 }