寻找以二进制字符串形式给出分数的游戏的赢家|第 2 集
原文:https://www . geeksforgeeks . org/find-一场比赛的获胜者-分数以二进制字符串集-2 给出/
给定一个代表排球比赛比分的二进制字符串 字符串。任务是根据以下条件找到比赛的获胜者:
- 在排球比赛中，两队互相比赛，先得 15 分的队将获胜，除非两队都得到了 14 分。
- 如果两队都得到了 14 分，那么保持领先 2 分的队将是胜者。
在给定的二进制字符串中， 0 表示极客队输一分， 1 表示极客队赢一分。任务是找出 GEEK 的团队是赢了还是输了这场比赛。
示例:
输入:str = " 01011111111111110110101 " T3】输出:GEEK s 赢了 T6】说明: GEEK 以 15-5 的比分获胜
输入:str = " 01010101010101010101010101010100 " 输出: GEEK 的输了 说明:对手以 16-14 的比分获胜
天真法:天真法在本题 集-1 中有提及。
高效进场:无论时间线是什么，最后得分的玩家都是赢家。原因如下:
游戏以某人赢得该盘并满足赢得游戏所需的条件而结束。
- 如果比赛结束时有人先得了 15 分，那么他将是最后一盘的赢家。
- 如果在达到 15-14 的状态后，有人保持两分领先赢得了比赛，那么最后一盘保持两分领先的人也将是获胜者。
下面是上述方法的实现。
C++
// C++ implementation of the above appoach
#include <bits/stdc++.h>
using namespace std;
// Function to find the winner
// from given timeline
string findTheWinner(string str, int N)
{
// If last point scored is 1 then
// GEEK is winner
if (str[N - 1] == '1') {
return "GEEK's won";
}
// Else GEEK lost
return "GEEK's lost";
}
// Driver Code
int main()
{
// Input score timeline
string str1 = "01011111111110110101";
int N1 = str1.size();
string str2 = "010101010101010101010101010100";
int N2 = str2.size();
// Print the winner
cout << findTheWinner(str1, N1) << endl;
cout << findTheWinner(str2, N2) << endl;
return 0;
}
Python 3
# Python code for the above approach
# Function to find the winner
# from given timeline
def findTheWinner(str, N):
# If last point scored is 1 then
# GEEK is winner
if (str[N - 1] == '1'):
return "GEEK's won"
# Else GEEK lost
return "GEEK's lost"
# Driver Code
# Input score timeline
str1 = "01011111111110110101"
N1 = len(str1)
str2 = "010101010101010101010101010100"
N2 = len(str2)
# Print the winner
print(findTheWinner(str1, N1))
print(findTheWinner(str2, N2))
# This code is contributed by Saurabh Jaiswal
C
// C# implementation of the above appoach
using System;
class GFG{
// Function to find the winner
// from given timeline
static string findTheWinner(string str, int N)
{
// If last point scored is 1 then
// GEEK is winner
if (str[N - 1] == '1')
{
return "GEEK's won";
}
// Else GEEK lost
return "GEEK's lost";
}
// Driver Code
public static void Main()
{
// Input score timeline
string str1 = "01011111111110110101";
int N1 = str1.Length;
string str2 = "010101010101010101010101010100";
int N2 = str2.Length;
// Print the winner
Console.WriteLine(findTheWinner(str1, N1));
Console.WriteLine(findTheWinner(str2, N2));
}
}
// This code is contributed by ukasp
java 描述语言
<script>
// JavaScript code for the above approach
// Function to find the winner
// from given timeline
function findTheWinner(str, N)
{
// If last point scored is 1 then
// GEEK is winner
if (str[N - 1] == '1') {
return "GEEK's won";
}
// Else GEEK lost
return "GEEK's lost";
}
// Driver Code
// Input score timeline
let str1 = "01011111111110110101";
let N1 = str1.length;
let str2 = "010101010101010101010101010100";
let N2 = str2.length;
// Print the winner
document.write(findTheWinner(str1, N1) + '<br>');
document.write(findTheWinner(str2, N2) + '<br>');
// This code is contributed by Potta Lokesh
</script>
Output
GEEK's won
GEEK's lost
时间复杂度: O(1) 辅助空间: O(1)
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