主函数如下: publicintreversePairs(int[]nums){intres=0;int[]copy=Arrays.copyOf(nums,nums.length);int[]bit=newint[copy.length+1];//对拷贝数组排序,为了得到大小排序过的位置。Arrays.sort(copy);for(intele:nums){//利用bit数据结构统计大于2*ele的元素个数res+=search(bit,index(copy,2L*ele+1)...
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)
https://leetcode.com/problems/reverse-pairs/discuss/97280/very-short-and-clear-mergesort-bst-java-solutions https://leetcode.com/problems/reverse-pairs/discuss/97268/general-principles-behind-problems-similar-to-reverse-pairs https://leetcode.com/problems/reverse-pairs/discuss/97311/evolve-from-brute...
Given an array nums, we call (i, j) an important reverse pair if i < j and nums[i] > 2*nums[j]. You need to return the number of important reverse pairs in the given array. Example1: Input: [1,3,2,3,1] Output: 2 Example2: Input: [2,4,3,5,1] Output: 3 Note: The ...
length); } private int reversePairsHelper (int[] nums, int[] copiedNums, int start, int end) { if (start + 1 == end) { return 0; } int middle = start + (end - start) / 2; int count = reversePairsHelper (nums, copiedNums, start, middle) + reversePairsHelper (nums, copied...
The problem you're dealing with, "Reverse Pairs" on LeetCode, involves counting the number of reverse pairs in an array. A reverse pair is defined as a pair (i, j) where i < j and nums[i] > 2 * nums[j]. Here are some test cases you should consider to ensure your solution wor...
2, 1), (4, 1), (4, 3) are reverse pairs. return 3 这道题跟LeetCode上的那道Count of Smaller Numbers After Self是一样的,唯一的一点点的小区别是那道题是返回一个向量,表示出原数组中每一个数字的右边比其小的数的个数,而这道题让我们求翻转对的总数,其实就是把每个数字右边比其小的数的个...
493. Reverse Pairs题目链接: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...
493. 翻转对 - 给定一个数组 nums ,如果 i < j 且 nums[i] > 2*nums[j] 我们就将 (i, j) 称作一个重要翻转对。 你需要返回给定数组中的重要翻转对的数量。 示例 1: 输入: [1,3,2,3,1] 输出: 2 示例 2: 输入: [2,4,3,5,1] 输出: 3 注意: 1. 给定数组的长度不会
Given n pairs of parentheses, write a function to generate all combinations of well-formed parenthes ... 【JAVA、C++】LeetCode 010 Regular Expression Matching Implement regular expression matching with support for '.' and '*'. '.' Matches any single character ... 【JAVA、C++】 LeetCode 008...