#91 Minimum Adjustment Cost

最新推荐文章于 2022-02-14 23:36:27 发布

原创最新推荐文章于 2022-02-14 23:36:27 发布 · 561 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#lintcode #dynamic programming

lintcode 专栏收录该内容

217 篇文章

订阅专栏

探讨如何通过调整数组中相邻整数间的差值不超过给定目标数，以最小化调整成本。利用动态规划方法，定义dp矩阵来求解最优解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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;
    }
};