lintcode: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]|

Have you met this question in a real interview? 

Yes

 Notice

You can assume each number in the array is a positive integer and not greater than 100.

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.

Tags 

转自http://www.cnblogs.com/yuzhangcmu/p/4153927.html

这道题真的惭愧，确实没有想出来

可以用递归和动态规划来解决这道题

递归，有点像word ladder和数独一样代码如下

/**     * @param A: An integer array.     * @param target: An integer.     */    public static int MinAdjustmentCost1(ArrayList<Integer> A, int target) {        // write your code here        if (A == null) {            return 0;        }                return rec(A, new ArrayList<Integer>(A), target, 0);    }        /*     * SOL 1:     * 最普通的递归方法。     * */    public static int rec(ArrayList<Integer> A, ArrayList<Integer> B, int target, int index) {        int len = A.size();        if (index >= len) {            // The index is out of range.            return 0;        }                int dif = 0;                int min = Integer.MAX_VALUE;                // If this is the first element, it can be from 1 to 100;        for (int i = 0; i <= 100; i++) {            if (index != 0 && Math.abs(i - B.get(index - 1)) > target) {                continue;            }                        B.set(index, i);            dif = Math.abs(i - A.get(index));            dif += rec(A, B, target, index + 1);            min = Math.min(min, dif);                        // 回溯            B.set(index, A.get(index));        }                return min;    }

更好的解法用动态规划，dp[i][[j] i表示现在loop到的位置，j表示1-100的数字，dp[i][j]表示，当loop到i位置时，取j值时整个修改的最小值

所以最终的结果要 比较   dp[A.size()-1][1--100]的大小来决定整个变换的最小值， 这个有点像刷颜色的问题

/*     * SOLUTION 4：     * DP     * */    /**     * @param A: An integer array.     * @param target: An integer.     */    public static int MinAdjustmentCost(ArrayList<Integer> A, int target) {        // write your code here        if (A == null || A.size() == 0) {            return 0;        }                // D[i][v]: 把index = i的值修改为v，所需要的最小花费        int[][] D = new int[A.size()][101];                int size = A.size();                for (int i = 0; i < size; i++) {            for (int j = 1; j <= 100; j++) {                D[i][j] = Integer.MAX_VALUE;                if (i == 0) {                    // The first element.                    D[i][j] = Math.abs(j - A.get(i));                } else {                    for (int k = 1; k <= 100; k++) {                        // 不符合条件                         if (Math.abs(j - k) > target) {                            continue;                        }                                                int dif = Math.abs(j - A.get(i)) + D[i - 1][k];                        D[i][j] = Math.min(D[i][j], dif);                    }                }            }        }                int ret = Integer.MAX_VALUE;        for (int i = 1; i <= 100; i++) {            ret = Math.min(ret, D[size - 1][i]);        }                return ret;    }

刷墙的问题：对比起来真是一模一样

class Solution {public:    /**     * @param costs n x 3 cost matrix     * @return an integer, the minimum cost to paint all houses     */    int minCost(vector<vector<int>>& costs) {        // Write your code here        if (costs.size() == 0) //检查边界条件非常重要            return 0;                int n = costs.size();                vector<vector<int>> dp(n, vector<int>(3));                //dp[i][j] 表示计算到第i户时，他选择第j种颜色图墙时的最小价钱                for (int i=0; i<3; i++)        {            dp[0][i] = costs[0][i];        }                        for (int i=1; i<n; i++)        {            for (int j=0; j<3; j++)            {                int minC = INT_MAX;                for (int k=0; k<3; k++)                {                    if (k != j)                    {                        minC = min(minC, dp[i-1][k]);                    }                    dp[i][j] = minC + costs[i][j];                }            }        }                int minRet = INT_MAX;        for (int i=0; i<3; i++)        {            minRet = min(minRet, dp[n-1][i]);        }                return minRet;    }};