最小调整代价-LintCode

给一个整数数组，调整每个数的大小，使得相邻的两个数的差小于一个给定的整数target，调整每个数的代价为调整前后的差的绝对值，求调整代价之和最小是多少。
样例:
对于数组[1, 4, 2, 3]和target=1，最小的调整方案是调整为[2, 3, 2, 3]，调整代价之和是2。返回2。
思想：动态规划

#ifndef C91_H
#define C91_H
#include
#include
#include
using namespace std;
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.empty() || A.size() < 2)
            return 0;
        int len = A.size();
        vector<vector<int>> dp(len, vector<int>(101, INT_MAX));//dp[i][j]存放A[i]变到j的最小调整代价
        for (int i = 0; i < 101; ++i)
            dp[0][i] = abs(i - A[0]);
        for (int i = 1; i < len; ++i)
        {
            for (int j = 0; j < 101; ++j)
            {
                //保证数字范围在0~100
                int upper = min(j + target, 100);
                int lower = max(j - target, 0);
                //将A[i]变到j,此时A[i-1]的取值为k,dp[i][j]为A[i-1]到k和A[i]到j的绝对值之和
                //dp[i][j]=min{dp[i-1][k]+abs(j-A[i])}
                for (int k = lower; k <= upper; ++k)
                    dp[i][j] = min(dp[i][j], dp[i - 1][k] + abs(j - A[i]));
            }
        }
        int res = INT_MAX;
        for (int i = 0; i < 101; ++i)
            res = min(res, dp[len - 1][i]);
        return res;
    }
};
#endif