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  • 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]| 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.

    这道题要看出是背包问题,不容易,跟FB一面 paint house很像,比那个难一点

    定义res[i][j] 表示前 i个number with 最后一个number是j,这样的minimum adjusting cost

    public int MinAdjustmentCost(ArrayList<Integer> A, int target) {
            // write your code here
            // 前i-1 个数调整后并且第i-1个数调整为j的cost
            int n = A.size();
            int[][] f = new int[n + 1][101];
            
            // initialize
            for (int i = 0; i <= 100; i++) {
                f[0][i] = 0;
            }
            for (int i = 1; i <= n; i++) {
                for (int j = 0; j <= 100; j++) {
                    f[i][j] = Integer.MAX_VALUE;
                }
            }
            //
            //function
            for (int i = 1; i <= n; i++) {
                for (int j = 0; j <= 100; j++) {
                    for (int k = 0; k <= 100; k++) {
                        if (Math.abs(j - k) <= target) {
                            
                                f[i][k] = Math.min(f[i][k], f[i - 1][j] + Math.abs(A.get(i - 1) - k));
                            
                            
                        }
                    }
                }
            }
            int ans = Integer.MAX_VALUE;
            for (int i = 0; i <= 100; i++) {
                ans = Math.min(ans, f[n][i]);
            }
            return ans;
        }
    

      

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  • 原文地址:https://www.cnblogs.com/apanda009/p/7294627.html
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