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| 1 | +想法:Divide and Conquer 的問題,我們可以先用每個 size 為 1 的格子作為葉節點;接著往上回傳時, |
| 2 | +如果上層節點的所有子節點都是葉節點且數值皆相同的話,我們就 free 掉原本記憶體並新增一個合併後的葉節點即可 |
| 3 | + |
| 4 | +Time Complexity : O(n^2) for T(n) = 4 * T(n / 2) + O(1) , by Master Theorem , T(n) = O(n^(log_2(4))) = O(n^2) |
| 5 | +(or you can see that every grids only been visited one times) |
| 6 | +Space Complexity : O(n^2) for the tree (recursion depth = O(logn)) |
| 7 | + |
| 8 | +/* |
| 9 | +// Definition for a QuadTree node. |
| 10 | +class Node { |
| 11 | +public: |
| 12 | + bool val; |
| 13 | + bool isLeaf; |
| 14 | + Node* topLeft; |
| 15 | + Node* topRight; |
| 16 | + Node* bottomLeft; |
| 17 | + Node* bottomRight; |
| 18 | + |
| 19 | + Node() { |
| 20 | + val = false; |
| 21 | + isLeaf = false; |
| 22 | + topLeft = NULL; |
| 23 | + topRight = NULL; |
| 24 | + bottomLeft = NULL; |
| 25 | + bottomRight = NULL; |
| 26 | + } |
| 27 | + |
| 28 | + Node(bool _val, bool _isLeaf) { |
| 29 | + val = _val; |
| 30 | + isLeaf = _isLeaf; |
| 31 | + topLeft = NULL; |
| 32 | + topRight = NULL; |
| 33 | + bottomLeft = NULL; |
| 34 | + bottomRight = NULL; |
| 35 | + } |
| 36 | + |
| 37 | + Node(bool _val, bool _isLeaf, Node* _topLeft, Node* _topRight, Node* _bottomLeft, Node* _bottomRight) { |
| 38 | + val = _val; |
| 39 | + isLeaf = _isLeaf; |
| 40 | + topLeft = _topLeft; |
| 41 | + topRight = _topRight; |
| 42 | + bottomLeft = _bottomLeft; |
| 43 | + bottomRight = _bottomRight; |
| 44 | + } |
| 45 | +}; |
| 46 | +*/ |
| 47 | + |
| 48 | +class Solution { |
| 49 | +public: |
| 50 | + Node* constructnode(vector<vector<int>>& grid,int size , int top , int left){ |
| 51 | + if (size == 1) |
| 52 | + return new Node( grid[top][left] ,true) ; |
| 53 | + |
| 54 | + Node* topleft = constructnode(grid , size / 2 , top , left) ; |
| 55 | + Node* topright = constructnode(grid , size / 2 , top , left + size/2) ; |
| 56 | + Node* bottomleft = constructnode(grid , size / 2 , top + size/2 , left); |
| 57 | + Node* bottomright = constructnode(grid , size/2, top+size/2,left + size/2); |
| 58 | + |
| 59 | + if (topleft->isLeaf && topright->isLeaf && |
| 60 | + bottomleft->isLeaf && bottomright->isLeaf && |
| 61 | + topleft->val == topright->val && |
| 62 | + topleft->val == bottomleft->val && |
| 63 | + topleft->val == bottomright->val) { |
| 64 | + delete(topleft) ; |
| 65 | + delete(topright) ; |
| 66 | + delete(bottomright) ; |
| 67 | + delete(bottomleft) ; |
| 68 | + return new Node(grid[top][left] , true) ; |
| 69 | + } |
| 70 | + return new Node(true , false , topleft,topright,bottomleft,bottomright) ; |
| 71 | + } |
| 72 | + Node* construct(vector<vector<int>>& grid) { |
| 73 | + return constructnode(grid , grid.size() , 0 , 0) ; |
| 74 | + } |
| 75 | +}; |
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