#91 Minimum Adjustment Cost

题目描述：

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]|

 Notice

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

100

.

Have you met this question in a real interview? 

Yes

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

.

题目思路：

这题的关键在于怎么去定义dp matrix。这里定义dp[i][j]，i表示A中index为i的那个数，j表示那个数的value。由于value是有限的(0 ~ 100)，所以对于每个i，我们都可以遍历一遍0~100去看这个cost各是多少。最后，由于最后一个数的value不确定，我们需要遍历dp[A.size() - 1][0...100]去找到最小cost的值。

Mycode（AC = 25ms）：

class Solution {
public:
/**
* @param A: An integer array.
* @param target: An integer.
*/
int MinAdjustmentCost(vector<int> A, int target) {
// write your code here
if (A.size() == 0) {
return 0;
}

vector<vector<int>> cost(A.size(), vector<int>(101, 0));

// cost[i][j] denotes the number at index i in A
// the number's value is j

// initialize the cost for first number (index i)
for (int i = 0; i <= 100; i++) {
cost[0][i] = abs(i - A[0]);
}

int min_cost = INT_MAX;

for (int i = 1; i < A.size(); i++) {

for (int j = 0; j <= 100; j++) {
cost[i][j] = INT_MAX;

// if number at index i and i - 1
// meets requirement, then find the min
// cost for all the possible i and i - 1
for (int k = 0; k <= 100; k++) {
if (abs(j - k) <= target) {
cost[i][j] = min(cost[i][j], cost[i - 1][k] + abs(A[i] - j));
}
}
}
}

// traverse through last number value of 0 ~ 100
// and find the min cost
for (int i = 0; i <= 100; i++) {
if (cost[A.size() - 1][i] < INT_MAX) {
min_cost = min(min_cost, cost[A.size() - 1][i]);
}
}

return min_cost;
}
};