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Well, a dynamic programming problem. Let's first define its statedp[i][j]to be the number of distinct subsequences oft[0..i - 1]ins[0..j - 1]. Then we... 阅读全文
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Well, the idea is to search from thetop-rightelement and then reduce the range for further searching by comparisons betweentargetand the current eleme... 阅读全文
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This problem is more or less the same asFind Minimum in Rotated Sorted Array. And one key difference is as stated in the solution tag. That is, due to... 阅读全文
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As explained in the Solution tag, the key to solving this problem is to use invariants. We set two pointers:lfor the left andrfor the right. One key i... 阅读全文
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For those who have already solvedSearch in Rotated Sorted Array, this problem can be solved similarly using codes for that problem and simply adding c... 阅读全文
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This problem is a nice application of binary search. The key lies in how to determine the correct half fortarget. Since the array has been rotated, we... 阅读全文
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Well, in the case of a linked list instead of an array, the problem becomes easier. We just need to find the node that will be the new head of the lis... 阅读全文
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This problem, as stated in the problem statement, has a lot of solutions. Since the problem requires us to solve it in O(1) space complexity, I only s... 阅读全文
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The first answer in this link has a nice illustrative explanation.Suppose the given array is like [nums[0], nums[1], nums[2], nums[3]]. Then the resul... 阅读全文
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As the note in the problem statement, this problem has a straight-forward O(n)-space solution, which is to generate the inorder traversal results of t... 阅读全文
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This problem has a nice BFS structure. Let's illustrate it using the examplenums = [2, 3, 1, 1, 4]in the problem statement. We are initially at positi... 阅读全文
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This problem has a very concise solution in this link, just 4-lines. The code is rewritten below.1 class Solution {2 public:3 bool canJump(vector&... 阅读全文
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Well, this problem is just a trick. In fact, we cannot really delete the given node, but just delete its next node after copying the data of the next ... 阅读全文
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Note: If you feel unwilling to read the long codes, just take the idea with you. The codes are unnecessarily long due to the inconvenient handle of ma... 阅读全文
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Note: The following idea is in fact from the last answer inthis link, which leads to a clean code. I just reorganize it and add some explanations. I h... 阅读全文