LeetCode: 975. Odd Even Jump 方国正 计算机硕士 来自专栏 · 进阶算法 题目描述:奇偶跳 给定一个整形数组A,从某一个下标开始,你可以进行一系列的跳跃。其中,第(1,3,5,...)次跳跃被称为奇数跳,而第(2,4,6,...)次跳跃被称为偶数跳。 你可以从下标 i 向前跳跃到下标 j ,其中 (i<j) 。本次...
https://github.com/grandyang/leetcode/issues/975 参考资料: https://leetcode.com/problems/odd-even-jump/ https://leetcode.com/problems/odd-even-jump/discuss/217974/Java-solution-DP-%2B-TreeMap https://leetcode.com/problems/odd-even-jump/discuss/217981/JavaC%2B%2BPython-DP-using-Map-or-S...
even[i]= odd[tm.get(floor)];if(odd[i]) ret++; tm.put(A[i],i);// as a result, it will always keep a biggest pos of A[i].}returnret; } }
代码参考花花。 /** @lc app=leetcodeid=975 lang=cpp** [975] Odd Even Jump** https://leetcode.com/problems/odd-even-jump/description/**algorithms* Hard (46.06%)* Likes: 409* Dislikes: 120* Total Accepted: 17.2K* Total Submissions: 37.4K* Testcase Example: '[10,13,12,14,15]'**...
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https://leetcode.com/problems/odd-even-jump/ 解题方法: DP: 由两个数组odd和even来代表在i这个位置上以odd jump和even jump,能不能跳到终点。 从后往前递推,最后一个点(终点)odd和even都是True. 在i点这个位置,如果能实现odd jump,那么odd[i] = even[此次odd jump的下一个点位置];如果能实现even ...
Leetcode Solutions (0x6A73). Contribute to waffle87/leetcode development by creating an account on GitHub.
The direction of a jump is determined as follows: if the point where the grasshopper was before the jump has aneven(偶数的)coordinate, the grasshopper jumps to the left, otherwise he jumps to the right. 移动方向由以下规则决定: 如果坐标是偶数,向左移 ...
对于每一个位置,无论是odd还是even jump,jump后的位置要么不存在,要么就是符合要求的唯一位置。因此最关键的问题是解决: 给定一个index i,找出jump后的位置 如果对每个index都能找到jump后的位置,那么利用递归 (DP) 很容易就能判断能否jump到最后一个位置。 由于数
During our 2nd jump (even numbered), we jump from i = 1 to i = 2 because A[2] is the largest value in (A[2], A[3], A[4]) that is less than or equal to A[1]. A[3] is also the largest value, but 2 is a smaller index, so we can only jump to i = 2 and not ...