技術標籤：leetcodeleetcode演算法java

leetcode學習筆記59

• 問題
• 思考
• • 方法1
  • 方法2

問題

Spiral Matrix II

Given a positive integer n, generate an n x n matrix filled with elements from 1 to n2 in spiral order.

Example 1:

Input: n = 3
Output: [[1,2,3],[8,9,4],[7,6,5]]

Example 2:

Input: n = 1
Output: [[1]]

Constraints

• 1 <= n <= 20

思考

轉方向就好了.

方法1

時間複雜度 O ( n 2 ) O(n^2) O(n2).
空間複雜度O(1).

class Solution {
    public int[][] generateMatrix(int n) {
        int left = 0, right = n - 1, top = 0, bottom = n - 1;
        int val = 1, max = n * n;
        int x = 0, y = -1;
        int[][] res = new int[n][n];
        while
 
(val <= max){
            
            y++;
            while(y <= right){
                res[x][y++] = val++;
                
            }
            top++;
            y--;
            
            x++;
            while(x <= bottom){
                res[x++][y] = val++;
            }
 

            right--;
            x--;
            
            y--;
            while(y >= left){
                res[x][y--] = val++;
            }
            bottom--;
            y++;
            
            x--;
            while(x >= top){
                res[x--][y] = val++;
            }
            left++;
            x++;

        }
        return res;
    }
}

方法2

更簡潔的寫法.

時間複雜度 O ( n 2 ) O(n^2) O(n2).
空間複雜度O(1).

class Solution {
    static int[][] moves = new int[][]{{0,1},{1,0},{0,-1},{-1,0}};
    
    public int[][] generateMatrix(int n) {
        int[][] res = new int[n][n];
        int cur = 1;
        for(int i = 0; i <= n/2; i++){
            
            for(int j = i; j < n-i; j++)
                res[i][j] = cur++;   
            
            for(int j = i+1; j < n-i; j++)
                res[j][n-i-1] = cur++;
            
            for(int j = n-i - 2; j >= i ; j--)
                res[n-i-1][j] = cur++;   
            
            for(int j = n-i-2; j > i ; j--)
                res[j][i] = cur++;
        }
        return res;
    }
}