标签归档:leetcode

Leetcode 685. Redundant Connection II(并查集)

不粘题面了。。。就是煞笔题写了半天,写出来各种细节错误。。
就是并查集判环,但因为是有向树只能固定方向了,不能按秩合并。而且删哪条边也需要考虑是否有一个点有两个父亲。然后就没了,讨论出细节即可。
这种状态准备面试还是csp都是gg的啊。。。

Leetcode 232. Implement Queue using Stacks(两个栈实现队列)

没什么卵用的东西,就是弄两个栈,再加入元素的时候辅助栈帮助实现逆序的过程,然后在加回去。。。

Leetcode 798. Smallest Rotation with Highest Score

Given an array A, we may rotate it by a non-negative integer K so that the array becomes A[K], A[K+1], A{K+2], … A[A.length – 1], A[0], A[1], …, A[K-1]. Afterward, any entries that are less than or equal to their index are worth 1 point.

For example, if we have [2, 4, 1, 3, 0], and we rotate by K = 2, it becomes [1, 3, 0, 2, 4]. This is worth 3 points because 1 > 0 [no points], 3 > 1 [no points], 0 <= 2 [one point], 2 <= 3 [one point], 4 <= 4 [one point]. Over all possible rotations, return the rotation index K that corresponds to the highest score we could receive. If there are multiple answers, return the smallest such index K. Example 1: Input: [2, 3, 1, 4, 0] Output: 3 Explanation: Scores for each K are listed below: K = 0, A = [2,3,1,4,0], score 2 K = 1, A = [3,1,4,0,2], score 3 K = 2, A = [1,4,0,2,3], score 3 K = 3, A = [4,0,2,3,1], score 4 K = 4, A = [0,2,3,1,4], score 3 So we should choose K = 3, which has the highest score. Example 2: Input: [1, 3, 0, 2, 4] Output: 0 Explanation: A will always have 3 points no matter how it shifts. So we will choose the smallest K, which is 0. Note: A will have length at most 20000. A[i] will be in the range [0, A.length]. 比赛的时候没想清楚以为要线段树觉得不可能那么麻烦就失去了梦想不想写了。。。 然后郑爹告诉我只需要树状数组就可以了。。 当a[i]<=i的时候,那么贡献的区间分为了两段,首先是从[1,i-a[i]+1],另外一段因为移动i+1次后a[i]变成了下标为n,所以就是[i+1,n]。 当a[i]>i的时候只有一段:[i,i+n-a[i]+1]。但是因为树状数组从1开始下标,而题目是0,所以整体加1在处理就好了。。。