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leetcode 390. Elimination Game | 390. 消除游戏（Java）

發(fā)布時間：2024/2/28 java 34 豆豆

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題目

https://leetcode.com/problems/elimination-game/

題解

方法1：模擬（內(nèi)存超出限制）

最樸素的思路，模擬每一輪的消除，直到剩余大小為 1 為止。

此方法內(nèi)存超限。

class Solution {public int lastRemaining(int n) {int[] a = new int[n];for (int i = 0; i < n; i++) {a[i] = i + 1;}int size = n;int[] b = null;boolean L = true; // start from leftwhile (size >= 1) {size /= 2;b = new int[size]; // 每輪縮容一半int p = 0;if (L) {while (p < size) {b[p] = a[p * 2 + 1];p++;}L = false;} else {while (p < size) {b[size - p - 1] = a[a.length - (p * 2 + 1) - 1];p++;}L = true;}int[] t = a;a = b;b = t;}return b[0];} }

方法2：只記錄起始位置以及步長（通過）

參考：JAVA: Easiest solution O(logN) with explanation

My idea is to update and record head in each turn. when the total number becomes 1, head is the only number left.

When will head be updated?

if we move from left
if we move from right and the total remaining number % 2 == 1
like 2 4 6 8 10, we move from 10, we will take out 10, 6 and 2, head is deleted and move to 4
like 2 4 6 8 10 12, we move from 12, we will take out 12, 8, 4, head is still remaining 2

then we find a rule to update our head.

example:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Let us start with head = 1, left = true, step = 1 (times 2 each turn), remaining = n(24)

we first move from left, we definitely need to move head to next position. (head = head + step)
So after first loop we will have:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 - > 2 4 6 8 10 12 14 16 18 20 22 24
head = 2, left = false, step = 1 * 2 = 2, remaining = remaining / 2 = 12

second loop, we move from right, in what situation we need to move head?
only if the remaining % 2 == 1, in this case we have 12 % 2 == 0, we don’t touch head.
so after this second loop we will have:
2 4 6 8 10 12 14 16 18 20 22 24 - > 2 6 10 14 18 22
head = 2, left = true, step = 2 * 2 = 4, remaining = remaining / 2 = 6

third loop, we move from left, move head to next position
after third loop we will have:
2 6 10 14 18 22 - > 6 14 22
head = 6, left = false, step = 4 * 2 = 8, remaining = remaining / 2 = 3

fourth loop, we move from right, NOTICE HERE:
we have remaining(3) % 2 == 1, so we know we need to move head to next position
after this loop, we will have
6 14 22 - > 14
head = 14, left = true, step = 8 * 2 = 16, remaining = remaining / 2 = 1

while loop end, return head

class Solution {public int lastRemaining(int n) {boolean L = true; // 是否從左向右int size = n;int step = 1;int begin = 1; // 開始的數(shù)字（而非位置）while (size > 1) {if (L || size % 2 == 1) {begin += step;}size /= 2;step *= 2;L = !L;}return begin;} }

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