Sword Finger Offer-47: The greatest value of gifts _ dynamic programming

Topic:

Each of the squares of a m*n board has a gift, each of which has a certain value (worth more than 0). You can start with the present in the box from the upper left corner of the chessboard and move one square at a time to the left or down, until you reach the lower right corner of the chessboard. Given a chessboard and the gift above it, calculate the most valuable gift you can take. Link:

Sword refers to offer (2nd edition): P233 Thinking Tags: algorithm: Recursive, dynamic Programming Solution: 1. Using the dynamic programming of the loop, the auxiliary two-dimensional array definition F (i,j) is used to represent the maximum value of the sum of gifts that can be obtained when a lattice arrives at coordinates (I,J). There are two ways to reach (I,J): (i-1,j) or (i,j-1); F (i,j) = Max (f (i-1,j), F (i , j-1)) + gift[i,j]; use loops to compute to avoid duplicate child problems.

 class Solution {public:int getmaxvalue_solution (const int* values, int rows, int cols) {if (values = = nullptr | | Rows <= 0 | |

        cols <= 0) return 0;
        int** maxvalues = new Int*[rows];

        for (int i = 0; i < rows; ++i) maxvalues[i] = new Int[cols];
                for (int i = 0; i < rows, ++i) {for (int j = 0; j < cols; ++j) {int left = 0;
                int up = 0;
                if (i > 0) up = maxvalues[i-1][j];

                if (J > 0) left = maxvalues[i][j-1];
            MAXVALUES[I][J] = Std::max (left, up) + Values[i*cols + j];

        an int maxValue = maxvalues[rows-1][cols-1];
        for (int i = 0; i < rows; ++i) delete[] maxvalues[i];

        Delete[] maxvalues;
    return maxValue; }
};

2. Optimization of space complexity, using one-dimensional array topics, it is known that the maximum gift value of coordinates (I,J) depends solely on coordinates (I-1,J) and (i,j-1) two lattices, so the maximum value above the i-2 line is not necessarily preserved. Use one-dimensional array to save: (0,0) ... (0,j-1) The preservation of (i,0) ... (i,j-1) The maximum value of (0,J) ... (0,cols-1) The preservation of (I-1,J) ... (I-1,cols-1) The greatest value. Each time a new (I,J) is computed, use the maximum value of the array subscript j-1 and J to add the current gift value.

Class Solution {public
:
    int getmaxvalue_solution (const int* values, int rows, int cols) {
        if (values = = NULLP TR | | Rows <= 0 | | cols <= 0) return
            0;

        int* maxvalues = new Int[cols];
        for (int i = 0; i < rows, ++i) {for
            (int j = 0; j < cols; ++j) {
                int left = 0;
                int up = 0;
                if (i > 0) up
                    = Maxvalues[j];
                if (J > 0) Left
                    = maxvalues[j-1];

                MAXVALUES[J] = Std::max (left, up) + Values[i*cols + j];
            }

        int maxValue = maxvalues[cols-1];

        Delete[] maxvalues;

        return maxValue;
    }
;