You have a total of n coins that you want to form in a staircase shape, where every k -th row must have exactly k coins.

Given n , find the total number of full staircase rows that can be formed.

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

Example 1:

n = 5

The coins can form the following rows:
¤
¤ ¤
¤ ¤

Because the 3rd row is incomplete, we return 2.

Example 2:

n = 8

The coins can form the following rows:
¤
¤ ¤
¤ ¤ ¤
¤ ¤

Because the 4th row is incomplete, we return 3.

这道题给了我们n个硬币，让我们按一定规律排列，第一行放1个，第二行放2个，以此类推，问我们有多少行能放满。通过分析题目中的例子可以得知最后一行只有两种情况，放满和没放满。由于是按等差数列排放的，我们可以快速计算出前i行的硬币总数。我们先来看一种O(n)的方法，非常简单粗暴，就是从第一行开始，一行一行的从n中减去，如果此时剩余的硬币没法满足下一行需要的硬币数了，我们之间返回当前行数即可，参见代码如下：

解法一：

class Solution {
public:
    int arrangeCoins(int n) {
        int cur = 1, rem = n - 1;
        while (rem >= cur + 1) {
            ++cur;
            rem -= cur;
        }
        return n == 0 ? 0 : cur;
    }
};

再来看一种O(lgn)的方法，用到了二分搜索法，我们搜索前i行之和刚好大于n的临界点，这样我们减一个就是能排满的行数，注意我们计算前i行之和有可能会整型溢出，所以我们需要将变量都定义成长整型，参见代码如下：

解法二：

class Solution {
public:
    int arrangeCoins(int n) {
        if (n <= 1) return n;
        long low = 1, high = n;
        while (low < high) {
            long mid = low + (high - low) / 2;
            if (mid * (mid + 1) / 2 <= n) low = mid + 1;
            else high = mid;
        }
        return low - 1;
    }
};

再来看一种数学解法O(1)，充分利用了等差数列的性质，我们建立等式, n = (1 + x) * x / 2, 我们用一元二次方程的求根公式可以得到 x = (-1 + sqrt(8 * n + 1)) / 2, 然后取整后就是能填满的行数，一行搞定简直丧心病狂啊：

解法三：

class Solution {
public:
    int arrangeCoins(int n) {
        return (int)((-1 + sqrt(1 + 8 * (long)n)) / 2);
    }
};

参考资料：

https://discuss.leetcode.com/topic/65664/my-55ms-c-solution

https://discuss.leetcode.com/topic/65575/java-o-1-solution-math-problem

https://discuss.leetcode.com/topic/65631/c-three-solutions-o-n-o-logn-o-1/2

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