README_EN.md

441. Arranging Coins

中文文档

Description

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

Solutions

Python3

class Solution:
    def arrangeCoins(self, n: int) -> int:
        return int(math.sqrt(2) * math.sqrt(n + 0.125) - 0.5)

class Solution:
    def arrangeCoins(self, n: int) -> int:
        left, right = 1, n
        while left < right:
            mid = (left + right + 1) >> 1
            s = ((1 + mid) * mid) >> 1
            if n < s:
                right = mid - 1
            else:
                left = mid
        return left

Java

class Solution {
    public int arrangeCoins(int n) {
        return (int) (Math.sqrt(2) * Math.sqrt(n + 0.125) - 0.5);
    }
}

class Solution {
    public int arrangeCoins(int n) {
        long left = 1, right = n;
        while (left < right) {
            long mid = (left + right + 1) >> 1;
            long s = ((1 + mid) * mid) >> 1;
            if (n < s) {
                right = mid - 1;
            } else {
                left = mid;
            }
        }
        return (int) left;
    }
}

C++

using LL = long;

class Solution {
public:
    int arrangeCoins(int n) {
        LL left = 1, right = n;
        while (left < right)
        {
            LL mid = left + right + 1 >> 1;
            LL s = (1 + mid) * mid >> 1;
            if (n < s) right = mid - 1;
            else left = mid;
        }
        return left;
    }
};

Go

func arrangeCoins(n int) int {
	left, right := 1, n
	for left < right {
		mid := (left + right + 1) >> 1
		s := (1 + mid) * mid >> 1
		if n < s {
			right = mid - 1
		} else {
			left = mid
		}
	}
	return left
}

...