标签:for cas nal while return com ati ber ant
Problem:
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.
Summary:
用n枚硬币摆成塔形,求可以摆成的完整的行数。
Analysis:
1.最简单的思路,依次减去递增的每行硬币数,直到n为非整数。
1 class Solution { 2 public: 3 int arrangeCoins(int n) { 4 int i = 0; 5 while (n > 0) { 6 i++; 7 n -= i; 8 } 9 10 return n == 0 ? i : i - 1; 11 } 12 };
2. 解一元二次方程:x^2 + x = 2 * n 解得:x = sqrt(2 * n + 1 / 4) - 1 /2
但要注意在此处n为32位有符号整型数,2 * n后有可能溢出,故在代码中应做相应处理。
1 class Solution { 2 public: 3 int arrangeCoins(int n) { 4 return sqrt((long long)2 * n + 0.25) - 0.5; 5 } 6 };
标签:for cas nal while return com ati ber ant
原文地址:http://www.cnblogs.com/VickyWang/p/6015456.html