2019/01/23

LeerCode 566. Reshape the Matrix

In MATLAB, there is a very useful function called 'reshape', which can reshape a matrix into a new one with different size but keep its original data.
You're given a matrix represented by a two-dimensional array, and two positiveintegers r and c representing the row number and column number of the wanted reshaped matrix, respectively.
The reshaped matrix need to be filled with all the elements of the original matrix in the same row-traversing order as they were.
If the 'reshape' operation with given parameters is possible and legal, output the new reshaped matrix; Otherwise, output the original matrix.
Example 1:
Input: 
nums = 
[[1,2],
 [3,4]]
r = 1, c = 4
Output: 
[[1,2,3,4]]
Explanation:
The row-traversing of nums is [1,2,3,4]. The new reshaped matrix is a 1 * 4 matrix, fill it row by row by using the previous list.
Example 2:
Input: 
nums = 
[[1,2],
 [3,4]]
r = 2, c = 4
Output: 
[[1,2],
 [3,4]]
Explanation:
There is no way to reshape a 2 * 2 matrix to a 2 * 4 matrix. So output the original matrix.
Note:
  1. The height and width of the given matrix is in range [1, 100].
  2. The given r and c are all positive.

這一題是要將一個原先是 r * c 的陣列轉換成 nr * nc 的陣列，沒有什麼特別的技巧，就是重新排列而已，因為是連續取值，所以原陣列的 row 跟 col 就用 ++ 的方式來處理，以加快速度。


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/**
 * Return an array of arrays of size *returnSize.
 * The sizes of the arrays are returned as *columnSizes array.
 * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
 */
int** matrixReshape(int** nums, int numsRowSize, int numsColSize, int r, int c, int** columnSizes, int* returnSize) {
    int **ret;
    int i, j, rr, cc;
    
    if (numsRowSize * numsColSize != r * c) {
        ret = nums;
        *returnSize = numsRowSize;
        columnSizes[0] = (int *) malloc (numsRowSize * sizeof(int));
        for (i=0; i<numsRowSize; i++) {
            columnSizes[0][i] = numsColSize;
        }
    } else {
        *returnSize = r;
        columnSizes[0] = (int *) malloc (r * sizeof(int));
        ret = (int **) malloc (r * sizeof(int *));
        rr = 0;
        cc = 0;
        for (i=0; i<r; i++) {
            ret[i] = (int *) malloc (c * sizeof(int));
            columnSizes[0][i] = c;
            for (j=0; j<c; j++) {
                ret[i][j] = nums[rr][cc];
                if (++cc == numsColSize) {
                    ++rr;
                    cc = 0;
                }
            }
        }
    }
    return ret;
}




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2019/01/19

leetcode 885. Spiral Matrix III

On a 2 dimensional grid with R rows and C columns, we start at (r0, c0)facing east.
Here, the north-west corner of the grid is at the first row and column, and the south-east corner of the grid is at the last row and column.
Now, we walk in a clockwise spiral shape to visit every position in this grid. 
Whenever we would move outside the boundary of the grid, we continue our walk outside the grid (but may return to the grid boundary later.) 
Eventually, we reach all R * C spaces of the grid.
Return a list of coordinates representing the positions of the grid in the order they were visited.

Example 1:
Input: R = 1, C = 4, r0 = 0, c0 = 0
Output: [[0,0],[0,1],[0,2],[0,3]]

Example 2:
Input: R = 5, C = 6, r0 = 1, c0 = 4
Output: [[1,4],[1,5],[2,5],[2,4],[2,3],[1,3],[0,3],[0,4],[0,5],[3,5],[3,4],[3,3],[3,2],[2,2],[1,2],[0,2],[4,5],[4,4],[4,3],[4,2],[4,1],[3,1],[2,1],[1,1],[0,1],[4,0],[3,0],[2,0],[1,0],[0,0]]

Note:
  1. 1 <= R <= 100
  2. 1 <= C <= 100
  3. 0 <= r0 < R
  4. 0 <= c0 < C

 這題用沒有向量的C不是很容易解，只好用暴力的方式求解，利用的方法很單純，就是讓 (r0, c0) 去繞圈圈，而且根據旋轉的特性，可以發現右下左上這樣的順序，格數變量是 1,1,2,2,3,3,4,4....，所以程式就長成下面這樣。leetcode 報 20ms，目前百分位置是 100%，期待出現更快的解法可以參考。

/**
 * Return an array of arrays of size *returnSize.
 * The sizes of the arrays are returned as *columnSizes array.
 * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
 */
int** spiralMatrixIII(int R, int C, int r0, int c0, int** columnSizes, int* returnSize) {
    int **ret;
    int i, j;
    int step;

    *returnSize = R * C;
    columnSizes[0] = (int *) malloc (R * C * sizeof(int));
    ret = (int **) malloc (R * C * sizeof(int *));

    i = 0;
    ret[i] = (int *) malloc (2 * sizeof(int));
    ret[i][0] = r0;
    ret[i][1] = c0;
    columnSizes[0][i] = 2;
    ++i;
    if (i == *returnSize) {
        return ret;
    }

    step = 1;
    while(1) {
        if (r0 >=0 && r0 < R) {
            for (j=1; j<=step; j++) {
                ++c0;
                if (c0 >= 0 && c0 < C) {
                    ret[i] = (int *) malloc (2 * sizeof(int));
                    ret[i][0] = r0;
                    ret[i][1] = c0;
                    columnSizes[0][i] = 2;
                    ++i;
                    if (i == *returnSize) {
                        return ret;
                    }
                }
            }
        } else {
            c0 += step;
        }
        if (c0 >= 0 && c0 < C) {
            for (j=1; j<=step; j++) {
                ++r0;
                if (r0 >=0 && r0 < R) {
                    ret[i] = (int *) malloc (2 * sizeof(int));
                    ret[i][0] = r0;
                    ret[i][1] = c0;
                    columnSizes[0][i] = 2;
                    ++i;
                    if (i == *returnSize) {
                        return ret;
                    }
                }
            }
        } else {
            r0 += step;
        }
        ++step;
        if (r0 >= 0 && r0 < R) {
            for (j=1; j<=step; j++) {
                --c0;
                if (c0 >= 0 && c0 < C) {
                    ret[i] = (int *) malloc (2 * sizeof(int));
                    ret[i][0] = r0;
                    ret[i][1] = c0;
                    columnSizes[0][i] = 2;
                    ++i;
                    if (i == *returnSize) {
                        return ret;
                    }
                }
            }
        } else {
            c0 -= step;
        }
        if (c0 >= 0 && c0 < C) {
            for (j=1; j<=step; j++) {
                --r0;
                if (r0 >=0 && r0 < R) {
                    ret[i] = (int *) malloc (2 * sizeof(int));
                    ret[i][0] = r0;
                    ret[i][1] = c0;
                    columnSizes[0][i] = 2;
                    ++i;
                    if (i == *returnSize) {
                        return ret;
                    }    
                }
            }
        } else {
            r0 -= step;
        }
        ++step;
    }
    return ret;
}
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