LeetCode 59. Spiral Matrix II

原题链接在这里：https://leetcode.com/problems/spiral-matrix-ii/

题目：

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

For example,
Given n = 3,

You should return the following matrix:

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

题解：

从外圈到内圈螺旋着填数，如果n是奇数的话，最后正中间的数需要单独填写.

Time Complexity: O(n^2).

Space: O(1), regardless res.

AC Java:

 1 class Solution {
 2     public int[][] generateMatrix(int n) {
 3         int [][] res = new int[n][n];
 4         int num = 1;
 5         int x = 0; 
 6         int y = 0;
 7         
 8         //由外到内，旋转加值
 9         while(n>1){
10             for(int j = 0; j<n-1; j++){
11                 res[x][y++] = num++;
12             }
13             
14             for(int i = 0; i<n-1; i++){
15                 res[x++][y] = num++;
16             }
17             
18             for(int j = 0; j<n-1; j++){
19                 res[x][y--] = num++;
20             }
21             
22             for(int i = 0; i<n-1; i++){
23                 res[x--][y] = num++;
24             }
25             
26             n-=2;
27             x++;
28             y++;
29         }
30         
31         if(n%2 == 1){
32             res[x][y] = num;
33         }
34         
35         return res;
36     }
37 }

Time Complexity: O(n ^ 2).

Space: O(1), regardless res.

AC Java:

 1 class Solution {
 2     public int[][] generateMatrix(int n) {
 3         int [][] res = new int[n][n];
 4         if(n < 1){
 5             return res;
 6         }
 7         
 8         int count = 1;
 9         int r1 = 0;
10         int r2 = n - 1;
11         int c1 = 0;
12         int c2 = n - 1;
13         while(r1 <= r2 && c1 <= c2){
14             for(int c = c1; c <= c2; c++){
15                 res[r1][c] = count++;
16             }
17             
18             for(int r = r1 + 1; r <= r2; r++){
19                 res[r][c2] = count++;
20             }
21             
22             if(r1 < r2 && c1 < c2){
23                 for(int c = c2 - 1; c >= c1; c--){
24                     res[r2][c] = count++;
25                 }
26                 
27                 for(int r = r2 - 1; r > r1; r--){
28                     res[r][c1] = count++;
29                 }
30             }
31             
32             r1++;
33             r2--;
34             c1++;
35             c2--;
36         }
37         
38         return res;
39     }
40 }

类似Spiral Matrix. 

跟上Spiral Matrix III.

posted @ 2015-09-26 04:12  Dylan_Java_NYC  阅读(384)  评论(0)  编辑  收藏  举报