[LintCode] Minimum Adjustment Cost

Minimum Adjustment Cost

Given an integer array, adjust each integers so that the difference of every adjacent 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]|

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

.

Note

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

100

.

SOLUTION 1:

这个题幸好是可以认为array里的数是不大于100的正整数，不然真不好做。

首先这题看起来问的是minimize the sum所以看起来应该相识DP。但是，DP又不好想转移方程，所以先用＊记忆化搜索＊去做这个。

也就DFS ＋ memory来做：

先来开一个result存放A[index] 变到[0-100]需要的cost，这里面大部分是不需要的，只留下跟上一位数相差target之内的数字。然后开一个helper(){}递归，然后接口需要输入的是，A数组，变化后的A数组，target，result，位置index。然后DFS的helper大致就是这样：

dfs(){

　　B.set(index, i);

　　dif = i - A.get(i);

　　dif = dif + dfs(...index + 1..);

　　result[index][i] = dif;

　　B.set(index, A.get(index));

}

然后加一个memory，在helper里面，就是result，记录每一个位置，每次的dif。

最后输出的结果是所有result里面的最小值。

代码：

public class Solution {
/**
* @param A: An integer array.
* @param target: An integer.
*/
//f(x,y) 把x转换到y距离为k之内的cost
//f(x,y) = min of (f(x - 1,y') + |A[x-1] - y|)
public int MinAdjustmentCost(ArrayList<Integer> A, int target) {
if (A == null || A.size() == 0){
return 0;
}
int size = A.size();
int[][] dp = new int[size][101];
for (int i = 0; i < size; i++){
for (int j = 1; j <= 100; j++){
dp[i][j] = Integer.MAX_VALUE;
if (i == 0){
dp[i][j] = Math.abs(j - A.get(i));
} else {
for (int k = 1; k <= 100; k++){
if (Math.abs(j - k) > target){
continue;
}
int dif = Math.abs(A.get(i) - j) + dp[i - 1][k];
dp[i][j] = Math.min(dp[i][j], dif);
}
}
}
}
int result = Integer.MAX_VALUE;
for (int i = 1; i <= 100; i++){
result = Math.min(result, dp[size - 1][i]);
}
return result;
}
}

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