Likou 59 - Spiral Matrix II - Boundary Judgment

Title description

Given you a positive integer n, generate a 1 to n2 all elements, and the elements are spirally arranged in clockwise order n x n Square matrix matrix .

Solution ideas

• We can create a new n*n empty matrix and add data to it according to the requirements of the title;
• Define the upper, lower, left, and right boundaries of the matrix as t , b , l , r ;
• The new num is initialized to 1 and incremented by 1 after each addition;
• Take a 3*3 matrix as an example: First, you need to fill the first row with 1,2,3 from left to right, so the loop condition at this time should be:

for(int i = l; i <= r; i++) arr[t][i] = num++; t++;

• The purpose of t++ is to move the upper bound (the blue line) down to shrink the matrix;
• Next need to add 4,5, the loop condition is:

for(int i = t; i <= b; i++) arr[i][r] = num++; r--;

• The green right border line moves to the left, and then fills 6,7, the loop condition is:

for(int i = r; i >= l; i--) arr[b][i] = num++; b--;

• The lower border line (red line) moves up, and then fills element 8, the loop condition is:

 for(int i = b; i >= t; i--) arr[i][l] = num++; l++;

• The left border line (yellow line) moves to the right, so far the cycle is over, and the outermost layer of elements has been filled;
• If the cycle is not finished, it will continue to repeat the above four steps and continuously adjust the four boundary lines;
• The condition for judging whether the loop is completed is num <= target. If num does not exceed target, it means that the element has not been filled.

Input and output example

Code

class Solution {public int[][] generateMatrix(int ​​n) {int num = 1, target = n*n;int l = 0, r = n-1, t = 0, b = n-1;int[][] arr = new int[n][n];while(num <= target){for(int i = l; i <= r; i++) arr[t][i] = num++;t++;for(int i = t; i <= b; i++) arr[i][r] = num++;r--;for(int i = r; i >= l; i--) arr[b][i] = num++;b--;for(int i = b; i >= t; i--) arr[i][l] = num++;l++;}return arr;}}