遍历每个状态 i ,然后遍历小于等于 i 的完全平方数 square ,则 i 可由 i - square 转移而来。 时间复杂度:O(n * sqrt(n)) 需要遍历 dp 中全部 O(n) 个状态,每次都需要遍历全部 O(sqrt(n)) 个完全平方数 空间复杂度:O(n) 需要维护 dp 中全部 O(n) 个状态 代码(Python3) class Solution: dp:
代码如下: 1classSolution(object):2defisPerfectSquare(self, num):3"""4:type num: int5:rtype: bool6"""7left, right =0, num8whileleft <=right:9mid = (left+right) / 210ifmid ** 2 ==num:11returnTrue12elifmid ** 2 <num:13left = mid + 114else:15right = mid -116returnFalse...
valid perfect square(leetcode) valid perfect square 题目介绍 解题方法 最近在刷leetcode, 遇到一个很easy的问题,但是用蛮力的方法,根本解不开,想着应该是数学得某个知识,这个结介绍一下。 题目介绍 这个题目就是输入一个整数,来判断它是不是一个整数的平方,虽然可以用蛮力算法来解决,但是在提价的时候会遇到...
Python3代码 classSolution:defnumSquares(self, n:int) ->int:# solution two: Lagrange's Four-square Theoremwhilen %4==0:# reduce nn /=4ifn %8==7:return4a =0whilea **2<= n: b =int((n - a **2) **0.5)ifa **2+ b **2== n:return(notnota) + (notnotb)# whether a an...
367. Valid Perfect Square aliblielite 来自专栏 · leetcode_python_easyclass Solution: def isPerfectSquare(self, num): """ :type num: int :rtype: bool """ # 法一 # return (int(num**0.5)**2)==num # # 法二:牛顿迭代法 # r = num #...
publicclassSolution{/** * @param n: a positive integer * @return: An integer */publicintnumSquares(intn) {// write your code hereint[] f = newint[n +1];for(inti =0; i <= n; i++) { f[i] = i;// max square number is i itself with all 1for(intj =1; j * j <= ...
[LeetCode]Perfect Squares 作者是 在线疯狂 发布于 2015年9月9日 在LeetCode. 题目描述: Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n. For example, given n = 12, return 3 because 12 = 4 + 4 + 4; ...
23 is a valid perfect square False Conclusion In this post, we saw one of the coding questions that are usually asked in interviews. Let us know how you would approach this problem in the comment section below. Happy coding! You may also like: ...
Complex Numbers in Python Complex numbers are of the form, ‘a + bj’, where a is real part floating value and b is the imaginary part floating value, and j represents the square root of −1. Example: 2.5 + 2j Number Type Conversion in Python There are a few built-in Python funct...
class Solution(object): def isPerfectSquare(self, num): """ :type num: int :rtype: bool """ for x in range(1000000): if