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Nothing to hard to think. Just take care of boundary conditions.class Solution {public: string convert(string s, int nRows) { if(s.empty() |... 阅读全文
posted @ 2014-07-23 12:54
Tonix
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Two passes: all pointer linking info will be recorded in 1st pass - to hashmap; 2nd pass recover the pointer relationship from hashmap.1A!class Soluti... 阅读全文
posted @ 2014-07-23 12:39
Tonix
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Another Double pointer solution. 1A!class Solution {public: ListNode *removeNthFromEnd(ListNode *head, int n) { ListNode *p = NULL; ... 阅读全文
posted @ 2014-07-23 11:41
Tonix
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2 solutions: bin-search and Newton iteration.class Solution {public: int _sqrt(long long tgt, long long i0, long long i1) { long long can... 阅读全文
posted @ 2014-07-23 05:41
Tonix
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Another textbook problem. We take back of postorder array as the current root, and then we can split the inorder array: 1st half for current right chi... 阅读全文
posted @ 2014-07-23 04:13
Tonix
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A collegiate textbook problem. Nothing special, but just take care of your memory use.class Solution {public: TreeNode *_buildTree(int pre[], int &... 阅读全文
posted @ 2014-07-23 03:29
Tonix
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