Leetcode: Spiral Matrix "54"

Leetcode: Spiral Matrix "54" title description

Given a matrix containing m x n elements (m rows, n columns), return all elements in the matrix in a clockwise spiral order.

Example 1:

Input: [[1, 2, 3], [4, 5, 6], [7, 8, 9]] output: [1,2,3,6,9,8,7,4,5]

Example 2:

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

Problem analysis

This problem is simply insane?! the way we use it is a lap of print !

The answer will be all elements in a clockwise order from the first outer layer, then the second outer element, and so on.

We first define four elements,R1,R2,C1,C2, which will frame a range, we print the value on the edge of the range clockwise, and then shrink the box again after each print .

Okay, is that a good question?

　　1. How many boxes to print?

Times=math.min (long, wide)%2==0? Math.min (long, wide)/2:math.min (long, wide)/2+1;

　　2. How do you change the horizontal axis of a clockwise print? Have color is the box to print

　　

3, spit Trough this question, simply disgusting.

Java

Class Solution {public    list<integer> spiralorder (int[][] matrix) {        list<integer> ans = new Arraylist<> ();        int m = matrix.length; Line        if (m = = 0)            return ans;        int n = matrix[0].length;//column        int c1  =  0;        int C2  = n-1;        int R1  = 0;        int r2 = m-1;        int times = Math.min (m,n)%2==0? Math.min (m,n)/2:math.min (m,n)/2+1;        for (int i=0;i<times;i++)        {for             (int c = c1; c <= C2; C + +) Ans.add (Matrix[r1][c]);            for (int r = r1 + 1; r <= R2; r++) Ans.add (Matrix[r][c2]);            if (R1 < R2 && C1 < C2) {for                (int c = c2-1; c > C1; c--) Ans.add (Matrix[r2][c]);                for (int r = r2; r > R1; r--) Ans.add (Matrix[r][c1]);            }            r1++;            r2--;            c1++;            c2--;        }        return ans;    }}

　　

Leetcode: Spiral Matrix "54"