原文链接: https://leetcode-cn.com/problems/find-the-winner-of-an-array-game

英文原文

Given an integer array arr of distinct integers and an integer k.

A game will be played between the first two elements of the array (i.e. arr[0] and arr[1]). In each round of the game, we compare arr[0] with arr[1], the larger integer wins and remains at position 0 and the smaller integer moves to the end of the array. The game ends when an integer wins k consecutive rounds.

Return the integer which will win the game.

It is guaranteed that there will be a winner of the game.

 

Example 1:

Input: arr = [2,1,3,5,4,6,7], k = 2
Output: 5
Explanation: Let's see the rounds of the game:
Round |       arr       | winner | win_count
  1   | [2,1,3,5,4,6,7] | 2      | 1
  2   | [2,3,5,4,6,7,1] | 3      | 1
  3   | [3,5,4,6,7,1,2] | 5      | 1
  4   | [5,4,6,7,1,2,3] | 5      | 2
So we can see that 4 rounds will be played and 5 is the winner because it wins 2 consecutive games.

Example 2:

Input: arr = [3,2,1], k = 10
Output: 3
Explanation: 3 will win the first 10 rounds consecutively.

Example 3:

Input: arr = [1,9,8,2,3,7,6,4,5], k = 7
Output: 9

Example 4:

Input: arr = [1,11,22,33,44,55,66,77,88,99], k = 1000000000
Output: 99

 

Constraints:

• 2 <= arr.length <= 105
• 1 <= arr[i] <= 106
• arr contains distinct integers.
• 1 <= k <= 109

中文题目

给你一个由 不同 整数组成的整数数组 arr 和一个整数 k 。

每回合游戏都在数组的前两个元素（即 arr[0] 和 arr[1] ）之间进行。比较 arr[0] 与 arr[1] 的大小，较大的整数将会取得这一回合的胜利并保留在位置 0 ，较小的整数移至数组的末尾。当一个整数赢得 k 个连续回合时，游戏结束，该整数就是比赛的 赢家 。

返回赢得比赛的整数。

题目数据 保证 游戏存在赢家。

 

示例 1：

输入：arr = [2,1,3,5,4,6,7], k = 2
输出：5
解释：一起看一下本场游戏每回合的情况：

因此将进行 4 回合比赛，其中 5 是赢家，因为它连胜 2 回合。

示例 2：

输入：arr = [3,2,1], k = 10
输出：3
解释：3 将会在前 10 个回合中连续获胜。

示例 3：

输入：arr = [1,9,8,2,3,7,6,4,5], k = 7
输出：9

示例 4：

输入：arr = [1,11,22,33,44,55,66,77,88,99], k = 1000000000
输出：99

 

提示：

• 2 <= arr.length <= 10^5
• 1 <= arr[i] <= 10^6
• arr 所含的整数 各不相同 。
• 1 <= k <= 10^9

通过代码

高赞题解

解题思路

就这么写呗。俩俩比较，并使得后面一个最大，不用移动操作

代码

class Solution {
public:
    int getWinner(vector<int>& arr, int k) {
        int i = 0, t = 0;  // t记录连胜次数
        while(t < k && i < arr.size() - 1){
            if(arr[i] > arr[i+1]){
                arr[i+1] = arr[i];     
                ++t;
            }else
                t = 1;
            ++i;
        }  
        return arr[i];
    }
};

统计信息

通过次数	提交次数	AC比率
13533	29814	45.4%

提交历史

提交时间	提交结果	执行时间	内存消耗	语言