54 spiral matrix

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Spiral Matrix

Description Title: Given a matrix of mxn elements (m row, n-column) comprising, follow a helical clockwise order, returns all the elements in the matrix.

示例 1:

输入:
[
 [ 1, 2, 3 ],
 [ 4, 5, 6 ],
 [ 7, 8, 9 ]
]
输出: [1,2,3,6,9,8,7,4,5]
示例 2:

输入:
[
  [1, 2, 3, 4],
  [5, 6, 7, 8],
  [9,10,11,12]
]
输出: [1,2,3,4,8,12,11,10,9,5,6,7]

Problem-solving ideas: hierarchical traversal operations from the outside in, the definition of the four directions, pay attention to border operation, this problem needs to do a few times, completely understood. Traversed in the FIG.

• For each layer, from the upper left in clockwise order through all the elements, assumed that the current layer is the top-left corner coordinates {(r1, c1)} (r1, c1), the lower right corner coordinates {(r2, c2)} ( r2, c2).
• First, through all the elements above {(r1, c)} (r1, c), according to {c = c1, ..., c2} c = c1, ..., the order of c2. Then traversing all the elements to the right of {(r, c2)} (r, c2), according {r = r1 + 1, ..., r2} r = r1 + 1, ..., r2 of the order. If this layer has four sides (i.e., {r1 <r2} r1 <r2 and {c1 <c2} c1 <c2), the following embodiment shown in FIG we traverse the elements of the left and below the element.
  

int* spiralOrder(int** matrix, int matrixSize, int* matrixColSize, int* returnSize){
    int r,c,r1,r2,c1,c2,count;
    if(!matrixSize||!matrixColSize[0]){
        *returnSize=0;
        return NULL;
    }
    int *result=(int *)malloc(sizeof(int)*matrixSize*matrixColSize[0]);
    *returnSize=matrixSize*matrixColSize[0];
    r1=c1=count=0;
    r2=matrixSize-1;
    c2=matrixColSize[0]-1;
    while(r1<=r2&&c1<=c2){
        for(c=c1;c<=c2;c++)result[count++]=matrix[r1][c];
        for(r=r1+1;r<=r2;r++)result[count++]=matrix[r][c2];
        if(r1<r2&&c1<c2){
            for(c=c2-1;c>c1;c--)result[count++]=matrix[r2][c];
            for(r=r2;r>r1;r--)result[count++]=matrix[r][c1];
        }
        r1++;
        r2--;
        c1++;
        c2--;
    }
    return result;
}