归并排序 MergeSort 和BIT可以解决,BST和 binary search不行 https://discuss.leetcode.com/topic/79227/general-principles-behind-problems-similar-to-reverse-pairsBST (binary search tree) BIT (binary indexed tree)
Binary Tree Postorder Traversal的C++提交中击败了0.88% 的用户逆波兰表达式求值 执行用时 : 32 ms, 在Evaluate Reverse Polish...leetcode栈 接雨水 简化路径 柱状图中最大的矩形 最大矩形 二叉树的中序遍历 二叉树的前序遍历 二叉树的锯齿形层次遍历 二叉树的后序遍历 逆波兰表达式求值 接雨水 给定 n 个非...
Binary Tree Upside Down Palindrome Linked List 参考资料: https://leetcode.com/problems/reverse-linked-list/ https://leetcode.com/problems/reverse-linked-list/discuss/58156/My-Java-recursive-solution https://leetcode.com/problems/reverse-linked-list/discuss/58337/Fast-Recursive-Java-solution https:/...
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Leetcode学习(24)—— Reverse Integer Reverse digits of an integer. Example1: x = 123, return 321 Example2: x = -123, return -321 1. 2. Note: The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows....
# write your code here left, right = 0, len(a) - 1 self.reverse_num = 0 self.merge_sort(a, left, right) return self.reverse_num def merge_sort(self, a, left, right): if left >= right: return mid = (left + right) // 2...
https://leetcode.com/problems... 和Count of Smaller Numbers After Self还有count of range sum是一类题,解法都差不多。BST可以做,但是这道题如果输入是有序的,简单的bst会超时,所以得用AVL来做。 然后就是binary index tree的做法,计算大于nums[j]2的时候就是拿全部的sum减去sum(nums[j]2) ...
⭐ Leetcode 解題紀錄 ⭐題型資料結構Python SolutionC++ SolutionNote ⭐BFS 相關題型 ⭐ 104 Maximum Depth of Binary Tree BFS (分層) Python 94 Binary Tree Inorder Traversal BFS (分層) Tree Python 內含 處理 Tree 樹問題的重點 102 Binary Tree Level Order Traversal BFS (分層) Tree Python ...
My Solutions to Leetcode problems. All solutions support C++ language, some support Java and Python. Multiple solutions will be given by most problems. Enjoy:) 我的Leetcode解答。所有的问题都支持C++语言,一部分问题支持Java语言。近乎所有问题都会提供多个算
package leetcode import "strings" func reverseWords151(s string) string { ss := strings.Fields(s) reverse151(&ss, 0, len(ss)-1) return strings.Join(ss, " ") } func reverse151(m *[]string, i int, j int) { for i <= j { (*m)[i], (*m)[j] = (*m)[j], (*m)[i]...