You are given two 0-indexed integer arrays player1
and player2
, that represent the number of pins that player 1 and player 2 hit in a bowling game, respectively.
The bowling game consists of n
turns, and the number of pins in each turn is exactly 10
.
Assume a player hit xi
pins in the ith
turn. The value of the ith
turn for the player is:
2xi
if the player hit10
pins in any of the previous two turns.- Otherwise, It is
xi
.
The score of the player is the sum of the values of their n
turns.
Return
1
if the score of player 1 is more than the score of player 2,2
if the score of player 2 is more than the score of player 1, and0
in case of a draw.
Example 1:
Input: player1 = [4,10,7,9], player2 = [6,5,2,3] Output: 1 Explanation: The score of player1 is 4 + 10 + 2*7 + 2*9 = 46. The score of player2 is 6 + 5 + 2 + 3 = 16. Score of player1 is more than the score of player2, so, player1 is the winner, and the answer is 1.
Example 2:
Input: player1 = [3,5,7,6], player2 = [8,10,10,2] Output: 2 Explanation: The score of player1 is 3 + 5 + 7 + 6 = 21. The score of player2 is 8 + 10 + 2*10 + 2*2 = 42. Score of player2 is more than the score of player1, so, player2 is the winner, and the answer is 2.
Example 3:
Input: player1 = [2,3], player2 = [4,1] Output: 0 Explanation: The score of player1 is 2 + 3 = 5 The score of player2 is 4 + 1 = 5 The score of player1 equals to the score of player2, so, there is a draw, and the answer is 0.
Constraints:
n == player1.length == player2.length
1 <= n <= 1000
0 <= player1[i], player2[i] <= 10
Solution: Simulation
We can write a function to compute the score of a player.
Time complexity: O(n)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: int isWinner(vector<int>& player1, vector<int>& player2) { auto score = [] (const vector<int>& player) { int sum = 0; int twice = 0; for (int x : player) { sum += twice-- > 0 ? x * 2 : x; if (x == 10) twice = 2; } return sum; }; int s1 = score(player1); int s2 = score(player2); if (s1 == s2) return 0; return s1 > s2 ? 1 : 2; } }; |
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