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 

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

這道題真的慚愧，確實沒有想出來

可以用遞歸和動態規劃來解決這道題

遞歸，有點像word ladder和數獨一樣代碼如下

/**
     * @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:
     * 最普通的遞歸方法。
     * */
    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;
    }

更好的解法用動態規劃，dp[i][[j] i表示現在loop到的位置，j表示1-100的數字，dp[i][j]表示，當loop到i位置時，取j值時整個修改的最小值

所以最終的結果要 比較   dp[A.size()-1][1--100]的大小來決定整個變換的最小值， 這個有點像刷顏色的問題

/*
     * 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的值修改爲v，所需要的最小花費
        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++) {
                        // 不符合條件 
                        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;
    }

刷牆的問題：對比起來真是一模一樣

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) //檢查邊界條件非常重要
            return 0;
        
        int n = costs.size();
        
        vector<vector<int>> dp(n, vector<int>(3));
        
        //dp[i][j] 表示計算到第i戶時，他選擇第j種顏色圖牆時的最小价錢
        
        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;
    }
};