LeetCode-螺旋矩阵

Posted 2021-03-18 zhouixi

思路：固定循环四条边，然后把四条边的数据清除形成

一个新的二维数组M*N代入即可。

第一次循环
[  
  [1, 2, 3, 4],
  [5, 6, 7, 8],
  [9,10,11,12],
  [13,14,15,16]

]

第二次循环就变成
[
　　 [6,7],
    [10,11]
]

当下一次循环的时候 ，数组变成长度为1直接add到list即可。无需循环

解：

class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
       
       List<Integer> list = new ArrayList();
        if (matrix.length == 0) {
            return list;
        }
        int len = Arrays.stream(matrix).mapToInt(arr -> arr.length).sum();

        //循环**
        /**
         *      [1, 2, 3, 4],
         *  *   [5, 6, 7, 8],
         *  *   [9,10,11,12],
         *  *   [13,14,15,16]
         *      {17,18,19,20}
         *
         */

        while (len != list.size()) {
            int[] arr = matrix[0];

            if(matrix.length == 1) {
                for(int i = 0 ; i < matrix.length ; i ++) {
                    int [] arrs = matrix[i];
                    Arrays.stream(arrs).forEach(item->{
                        list.add(item);
                    });
                }
                break;
            }
            
            for(int i = 0 ; i < matrix.length; i ++) {
                int [] arrs = matrix[i];
                if(arrs.length == 1) {
                    list.add(arrs[0]);
                }
            }
            if(list.size() == len) {
                break;
            }

            int [][] newArr = new int[matrix.length-1][];
            //第一次横排
            for (int k = 0; k < arr.length; k++) {
                list.add(arr[k]);
            }
            for(int i = 0 ; i < newArr.length; i ++) {
                newArr[i] = matrix[i+1];
            }

            //第一次纵排
            for(int y = 0 ; y < matrix.length ; y ++ ) {
                int [] colArr = matrix[y];
                if(y == 0) {
                    continue;
                }
               list.add(colArr[colArr.length-1]);
            }
            int [][] newArr1 = new int [newArr.length][];
            for(int i = 0 ; i < newArr.length; i ++ ) {
                int [] horArr = new int[newArr[0].length-1];
                for(int k = 0 ; k < horArr.length ; k ++ ) {
                    horArr[k] = newArr[i][k];
                }
                newArr1[i] = horArr;
            }

            //第二次横排
            int [] lastArr = matrix[matrix.length-1];
            for(int z = lastArr.length - 1 ; z >= 0 ; z--) {
                if(z == lastArr.length - 1) {
                    continue;
                }
                list.add(lastArr[z]);
            }
            int [][] newArr2 = new int [newArr1.length-1][];
            for(int z = 0 ; z < newArr1.length - 1 ; z++) {
                int [] arr2 = newArr1[z];
                newArr2[z] = arr2;
            }

            //第二次纵排
            for(int i = matrix.length - 1 ; i >= 0 ; i -- ) {
                int [] colArr = matrix[i];
                if(i == 0 || i == matrix.length -1 ) {
                    continue;
                }
                list.add(colArr[0]);
            }
            int [][] newArr3 = new int [newArr2.length][];
            for(int i = 0 ; i < newArr3.length; i ++ ) {
                int [] horArr = new int[newArr2[0].length-1];
                for(int k = 0 ; k < horArr.length ; k ++ ) {
                    horArr[k] = newArr2[i][k+1];
                }
                newArr3[i] = horArr;
            }
            matrix = newArr3 ;
        }
        return list;
    }
}