Given an array with n integers, you need to find if there are triplets (i, j, k) which satisfies following conditions: 0 < i, i + 1 < j, j + 1 < k < n - 1 Sum of subarrays (0, i - 1), (i + 1, j - 1), (j + 1, k - 1) and (k + 1, n - 1) should be…
Given an array with n integers, you need to find if there are triplets (i, j, k) which satisfies following conditions: 0 < i, i + 1 < j, j + 1 < k < n - 1 Sum of subarrays (0, i - 1), (i + 1, j - 1), (j + 1, k - 1) and (k + 1, n - 1) should be…
Given an array with n integers, you need to find if there are triplets (i, j, k) which satisfies following conditions: 0 < i, i + 1 < j, j + 1 < k < n - 1 Sum of subarrays (0, i - 1), (i + 1, j - 1), (j + 1, k - 1) and (k + 1, n - 1) should be…
作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 暴力 日期 题目地址:https://leetcode-cn.com/problems/split-array-with-equal-sum/ 题目描述 Given an array with n integers, you need to find if there are triplets (i, j, k) which satisfies fol…
In a given integer array A, we must move every element of A to either list B or list C. (B and C initially start empty.) Return true if and only if after such a move, it is possible that the average value of B is equal to the average value of C, and…
[题目链接]:https://leetcode.com/contest/leetcode-weekly-contest-26/problems/split-array-with-equal-sum/ [题意] 让你把一段序列去掉3个元素,然后分成4个部分; 要求这4个部分的和相同; 问你可不可能; [题解] 先枚举要删除的3个元素中的中间那个元素j; 然后把整个序列分成左边和右边两个部分; 然后再左边的序列中枚举i; 然后把0..j-1分成0..i-1和i+1..j-1两个部分; 然后看这两个部…
You are given an integer array sorted in ascending order (may contain duplicates), you need to split them into several subsequences, where each subsequences consist of at least 3 consecutive integers. Return whether you can make such a split. Example…
Given a string S of digits, such as S = "123456579", we can split it into a Fibonacci-like sequence [123, 456, 579]. Formally, a Fibonacci-like sequence is a list F of non-negative integers such that: 0 <= F[i] <= 2^31 - 1, (that is, each…
思路: 使用哈希表降低复杂度.具体来说: 枚举j: 枚举i,如果sum[i - 1] == sum[j - 1] - sum[i],就用哈希表把sum[i - 1]记录下来: 枚举k,如果sum[k - 1] - sum[j] == sum[n - 1] - sum[k]并且哈希表中存在sum[k - 1] - sum[j],返回true. 返回false. 实现: class Solution { public: bool splitArray(vector<int>& nums)…
Given an array A, partition it into two (contiguous) subarrays left and right so that: Every element in left is less than or equal to every element in right. left and right are non-empty. left has the smallest possible size. Return the length of left…