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| 1 | +想法:先走過一次 list 算出總共長度 size , 實際上 k 等價於 k % size 的 rotate 次數; |
| 2 | +因為每次 rotate 是把最後一個元素放到最前面,所以 k 次 rotate 等價於 n - k 次把最前面的元素放到最後 |
| 3 | +因此,我們只需執行 n - k 次將頭部放到尾部的操作即可 |
| 4 | + |
| 5 | +// assume there are n nodes in list |
| 6 | +Time Complexity : O(n) for traversing the list |
| 7 | +Space Complexity : O(1) for variables |
| 8 | + |
| 9 | +/** |
| 10 | + * Definition for singly-linked list. |
| 11 | + * struct ListNode { |
| 12 | + * int val; |
| 13 | + * ListNode *next; |
| 14 | + * ListNode() : val(0), next(nullptr) {} |
| 15 | + * ListNode(int x) : val(x), next(nullptr) {} |
| 16 | + * ListNode(int x, ListNode *next) : val(x), next(next) {} |
| 17 | + * }; |
| 18 | + */ |
| 19 | +class Solution { |
| 20 | +public: |
| 21 | + ListNode* rotateRight(ListNode* head, int k) { |
| 22 | + if ( !head ) |
| 23 | + return nullptr ; |
| 24 | + |
| 25 | + ListNode* cursor = head , *tail ; |
| 26 | + int size = 0 ; |
| 27 | + |
| 28 | + while (cursor) { |
| 29 | + size++ ; |
| 30 | + tail = cursor ; |
| 31 | + cursor = cursor->next; |
| 32 | + } |
| 33 | + |
| 34 | + k %= size ; |
| 35 | + if (k == 0) |
| 36 | + return head ; |
| 37 | + k = size - k ; |
| 38 | + while ( k-- ) { |
| 39 | + ListNode* newtail = head , *newhead = head->next ; |
| 40 | + tail->next = newtail ; |
| 41 | + newtail->next = nullptr ; |
| 42 | + tail = newtail ; |
| 43 | + head = newhead ; |
| 44 | + } |
| 45 | + return head; |
| 46 | + } |
| 47 | +}; |
| 48 | + |
| 49 | +// 法二:一樣先找出等價的 k <= n ,接著把尾部與頭部相連,再把第 n - k - 1 個(0-index)元素設成尾部 |
| 50 | +輸出第n - k 個元素即可 |
| 51 | +/** |
| 52 | + * Definition for singly-linked list. |
| 53 | + * struct ListNode { |
| 54 | + * int val; |
| 55 | + * ListNode *next; |
| 56 | + * ListNode() : val(0), next(nullptr) {} |
| 57 | + * ListNode(int x) : val(x), next(nullptr) {} |
| 58 | + * ListNode(int x, ListNode *next) : val(x), next(next) {} |
| 59 | + * }; |
| 60 | + */ |
| 61 | +class Solution { |
| 62 | +public: |
| 63 | + ListNode* rotateRight(ListNode* head, int k) { |
| 64 | + if ( !head ) |
| 65 | + return nullptr ; |
| 66 | + |
| 67 | + ListNode* cursor = head , *tail ; |
| 68 | + int size = 0 ; |
| 69 | + |
| 70 | + while (cursor) { |
| 71 | + size++ ; |
| 72 | + tail = cursor ; |
| 73 | + cursor = cursor->next; |
| 74 | + } |
| 75 | + |
| 76 | + k %= size ; |
| 77 | + if (k == 0) |
| 78 | + return head ; |
| 79 | + tail->next = head ; |
| 80 | + cursor = head ; |
| 81 | + k = size - k ; |
| 82 | + while (--k) { |
| 83 | + cursor = cursor->next ; |
| 84 | + } |
| 85 | + ListNode* newhead = cursor->next ; |
| 86 | + cursor->next = nullptr; |
| 87 | + return newhead ; |
| 88 | + } |
| 89 | +}; |
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