LeetCode-441 - Arranging coins

Arrange coins

Problem description: You have a total of n coins, you need to arrange them in a ladder shape, and there must be exactly k coins in the kth row.

Given a number n, find the total number of rows that form a complete ladder row.

n is a non-negative integer and is in the range of a 32-bit signed integer.

For example descriptions, please refer to the official website of LeetCode.

Source: LeetCode
Link: https://leetcode-cn.com/problems/arranging-coins/
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Solution 1: Exhaustive method

Simply accumulate until it is greater than n, and finally return the corresponding number of layers. This method is too inefficient and will time out when n is large.

Solution 2: Binary search method

First, the upper and lower limits low and high are the maximum and minimum layers respectively. The maximum value is estimated according to n = x * (x + 1) / 2 , and then use the binary search method to find the maximum number of layers that can be placed, and finally return the number of layers.

 public class LeetCode_441 {
    /**
     * 穷举法，n超大时会超时，效率不高
     *
     * @param n
     * @return
     */
    public static int arrangeCoins(int n) {
        int sum = 0, rows = 0;
        for (int i = 1; ; i++) {
            if (sum + i > n) {
                break;
            }
            sum += i;
            rows++;
        }
        return rows;
    }

    /**
     * 二分查找法
     *
     * @param n
     * @return
     */
    public static int arrangeCoins2(int n) {
        // low和high分别是最大和最小的层数，最大值根据 `n = x * (x + 1) / 2` 估算得到
        int low = 1, high = (int) Math.sqrt(Double.valueOf(Integer.MAX_VALUE) * 2), mid = -1;
        // 利用二分查找法找到最多可以放到第几层
        while (low <= high) {
            mid = (low + high) / 2;
            double temp = (double) mid * (mid + 1) / 2;
            if (temp > n) {
                high = mid - 1;
            } else if (temp < n) {
                low = mid + 1;
            } else {
                return mid;
            }
        }
        double temp = (double) mid * (mid + 1) / 2;
        if (temp > n) {
            return mid - 1;
        } else {
            return mid;
        }
    }

    public static void main(String[] args) {
        // 测试用例一，期望输出： 2
        System.out.println(arrangeCoins2(5));
        // 测试用例二，期望输出： 3
        System.out.println(arrangeCoins2(8));
        // 测试用例三，期望输出：65535
        System.out.println(arrangeCoins2(2147483647));
        // 测试用例三，期望输出：60070
        System.out.println(arrangeCoins2(1804289383));
    }
}

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