7/9 每日一題(等比級數, 二分搜索法)

You have n coins and you want to build a staircase with these coins. The staircase consists of k rows where the ith row has exactly i coins. The last row of the staircase may be incomplete.

Given the integer n, return the number of complete rows of the staircase you will build.

 

Example 1:

Input: n = 5
Output: 2
Explanation: Because the 3rd row is incomplete, we return 2.

Example 2:

Input: n = 8
Output: 3
Explanation: Because the 4th row is incomplete, we return 3.

 

Constraints:

• 1 <= n <= 231 - 1

class Solution:

    def arrangeCoins(self, n: int) -> int:

        floor=0

        floor_sum = 0

        while floor_sum <= n:

            floor +=1

            floor_sum +=floor            

        if floor_sum==n:

            return floor

        else:

            return floor-1

------------------------參考解答 二分法

def f(n):

    left,right=0,n

    while left <= right:

        mid=left+(right-left)//2

        total=((mid+1)*mid)//2 #計算中間行所需的總硬幣

        #首相是1,公差是1一直加到mid項

        #例如mid=3 , 3*4//2=6 表示地3行需要6個硬幣

        if total == n:

            return print(f'找到mid={mid},total={total}')

        elif total > n:

            print(f'left={left}')

            print(f'right={right}')            

            print(f'mid={mid}')

            print(f'total={total}')

            right=mid -1

            print(f'mid超出n,right左移={right}')

            print()

        else:

            print(f'left={left}')

            print(f'right={right}')            

            print(f'mid={mid}')

            print(f'total={total}')

            left = mid+1

            print(f'mid小於n,left右移={left}')

            print()

    return print(f'返回最大完整數={right}')

f(5)

---------------------

left=0 right=10 mid=5 total=15 mid超出n,right左移=5 left=0 right=5 mid=2 total=3 mid小於n,left右移=3 找到mid=4,total=10