摘要:It is a matter of hashmap use. For the follow-up:
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01 2017 档案
摘要:A more programming-like solution, is to hack the problem from simple: we try each possble base value, and see which 111..11 fits target number - using
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摘要:Check out this brilliant solution:https://discuss.leetcode.com/topic/70658/concise-java-solution-using-dp My first solution was a O(n^2) typical DP bo
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摘要:The punch line of this one: sum of leaves: pi..pj = root..pj - root..pi.
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摘要:Split it into 2 group of arrays, 2 arrays in each group. Then use hashmap. class Solution { public: int fourSumCount(vector<int>& A, vector<int>& B, v
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摘要:A natural stack based solution. Seriously, whey 'Medium'? /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; *
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摘要:Check this out: https://discuss.leetcode.com/topic/76312/java-1-line-recursion-solution Neat thoughts with neat code. Bravo. Deserve to learn.
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摘要:Simple greedy thought .. satify each kid with minimum qualified cookie, from the least greedy kid..
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摘要:Idea is, among all factors of the int, we pick the two that is the closest pair. And searching from sqrt(area) is a better idea: https://discuss.leetc
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摘要:I prefer shorter code, so this one:https://discuss.leetcode.com/topic/75448/memoization-c-solution Please note: we don't want to be too greedy - despe
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摘要:Yes the most concise solution is recursion.https://discuss.leetcode.com/topic/66165/very-concise-c-solution-for-general-binary-tree-not-only-bst Here
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摘要:Natural Greedy. Sort the array and pick the medium as target.
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摘要:Fun one. A matter of string generation by given rules. I know it can be much shorter.. but i'm lazy to do that.
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摘要:Mediocre solution is O(nlgn) by using max-heap - but, remember bucket sort? it is O(n)
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摘要:This problem reflects a key strategy when dealing with discrete problem space - split split split.. That' is to say, we check each house and find its
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摘要:Brutal force works - and a smarter way is to utilize KMP. The most popular two solutions are essentially the same - one is obvious KMP, and another on
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摘要:it is all about corner-cases...
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摘要:A trickier DFS, with a little bit complex recursion param tweak, and what's more important is pruning strategies.. or your code will TLE.
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摘要:Fun one.. the punch line of this problem is quite common in Bit related problems on HackerRank - visualize it in your mind, and you will find: all bit
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摘要:My first reaction is to have an unlimited length of bit-array, to mark existence. But if no extra mem is allowed, we can simply use 'sign' on each ind
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摘要:The follow-up question is fun: "Could you do it in one-pass, using only O(1) extra memory and without modifying the value of the board?" When we meet
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