LeetCode,leetcodeoj
題目連結:Spiral Matrix II
Given an integer n, generate a square matrix filled with elements from 1 to n^2 in spiral order.
For example,
Given n = 3,
You should return the following matrix:
[ [ 1, 2, 3 ], [ 8, 9, 4 ], [ 7, 6, 5 ] ]
這道題的要求是返回長寬均為n的矩陣,其元素是按照1~n^2的螺旋順序排列。
和Spiral Matrix同樣簡單的數組操作問題,只需要按右、下、左、上的順序逐行或列遍曆數組。不過在處理邊界問題上,這題貌似更容易一些:可以先初始化二維數組均為0,然後填寫的時候碰到非0值的時候就改變方向即可。
時間複雜度:O(n2)
空間複雜度:O(n2)
1 class Solution 2 { 3 public: 4 vector<vector<int> > generateMatrix(int n) 5 { 6 vector<vector<int> > vvi(n, vector<int>(n, 0)); 7 8 if(n < 1) 9 return vvi;10 11 int i = 0, j = 0, k = 1;12 vvi[i][j] = k;13 while(k < n * n)14 {15 while(j + 1 < n && vvi[i][j + 1]==0)16 vvi[i][++ j] = ++ k;17 18 while(i + 1 < n && vvi[i + 1][j]==0)19 vvi[++ i][j] = ++ k;20 21 while(j - 1 >= 0 && vvi[i][j - 1]==0)22 vvi[i][-- j] = ++ k;23 24 while(i - 1 >= 0 && vvi[i - 1][j]==0)25 vvi[-- i][j] = ++ k;26 }27 28 return vvi;29 }30 };
當然,由於這裡處理邊界問題比較統一,因此也可以將四個方向的移動合并到一起,通過move = [[0, 1], [1, 0], [0, -1], [-1, 0]]數組控制移動。
1 class Solution 2 { 3 public: 4 vector<vector<int> > generateMatrix(int n) 5 { 6 vector<vector<int> > vvi(n, vector<int>(n, 0)); 7 8 if(n < 1) 9 return vvi;10 11 int move[4][2] = { {0, 1}, {1, 0}, {0, -1}, {-1, 0} };12 13 int x = 0, y = 0, k = 1;14 vvi[0][0] = k;15 while(k < n * n)16 for(int i = 0; i < 4; ++ i)17 while(x + move[i][0] >= 0 && x + move[i][0] < n &&18 y + move[i][1] >= 0 && y + move[i][1] < n &&19 vvi[x + move[i][0]][y + move[i][1]] == 0)20 vvi[x += move[i][0]][y += move[i][1]] = ++ k;21 22 return vvi;23 }24 };
轉載請說明出處:LeetCode --- 59. Spiral Matrix II