From 2e2934db95472a1e1ff661f083c92691e4cc5b9c Mon Sep 17 00:00:00 2001 From: halfrost Date: Thu, 20 Jan 2022 22:22:22 +0800 Subject: [PATCH] Add solution 390 --- .../0300~0399/0390.Elimination-Game.md | 81 +++++++++++++++++++ 1 file changed, 81 insertions(+) create mode 100644 website/content/ChapterFour/0300~0399/0390.Elimination-Game.md diff --git a/website/content/ChapterFour/0300~0399/0390.Elimination-Game.md b/website/content/ChapterFour/0300~0399/0390.Elimination-Game.md new file mode 100644 index 00000000..d90aca5b --- /dev/null +++ b/website/content/ChapterFour/0300~0399/0390.Elimination-Game.md @@ -0,0 +1,81 @@ +# [390. Elimination Game](https://leetcode.com/problems/elimination-game/) + + +## 题目 + +You have a list `arr` of all integers in the range `[1, n]` sorted in a strictly increasing order. Apply the following algorithm on `arr`: + +- Starting from left to right, remove the first number and every other number afterward until you reach the end of the list. +- Repeat the previous step again, but this time from right to left, remove the rightmost number and every other number from the remaining numbers. +- Keep repeating the steps again, alternating left to right and right to left, until a single number remains. + +Given the integer `n`, return *the last number that remains in* `arr`. + +**Example 1:** + +``` +Input: n = 9 +Output: 6 +Explanation: +arr = [1, 2,3, 4,5, 6,7, 8,9] +arr = [2,4, 6,8] +arr = [2, 6] +arr = [6] + +``` + +**Example 2:** + +``` +Input: n = 1 +Output: 1 + +``` + +**Constraints:** + +- `1 <= n <= 109` + +## 题目大意 + +列表 arr 由在范围 [1, n] 中的所有整数组成,并按严格递增排序。请你对 arr 应用下述算法: + +- 从左到右,删除第一个数字,然后每隔一个数字删除一个,直到到达列表末尾。 +- 重复上面的步骤,但这次是从右到左。也就是,删除最右侧的数字,然后剩下的数字每隔一个删除一个。 +- 不断重复这两步,从左到右和从右到左交替进行,直到只剩下一个数字。 + +给你整数 n ,返回 arr 最后剩下的数字。 + +## 解题思路 + +- 模拟题。按照题意,第一轮从左往右删除数字,第二轮从右往左删除数字。题目要求最后剩下的数字,模拟过程中不需要真的删除元素。只需要标记起始元素,该轮步长和方向即可。最后总元素只剩下一个即为所求。 + +## 代码 + +```go +package leetcode + +func lastRemaining(n int) int { + start, dir, step := 1, true, 1 + for n > 1 { + if dir { // 正向 + start += step + } else { // 反向 + if n%2 == 1 { + start += step + } + } + dir = !dir + n >>= 1 + step <<= 1 + } + return start +} +``` + + +---------------------------------------------- +
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