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[Java教程]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 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.

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 }