classSolution {public:booljudgeSquareSum(intc) {longa =0, b =sqrt(c);while(a <=b) {if(a * a + b * b == c)returntrue;elseif(a * a + b * b < c) ++a;else--b; }returnfalse; } }; 下面这种解法基于费马平方和定理 Fermat's theorem
费马平方和定理 Fermat's theorem on sums of two squares的一般推广形式:当某个数字的 4k+3 型的质数因子的个数均为偶数时,其可以拆分为两个平方数之和(each prime that is congruent to 3 mod 4 appears with an even exponent in the prime factorization of the number)。那么我们只要统计其质数因子的个...
从0 和sqrt(n)两端分别查找 public boolean judgeSquareSum2(int c) { int low = 0; int high = (int)Math.sqrt(c); while(low<=high) { int sum = low*low+high*high; if(sum<c) { low++; } else if(sum>c) { high--; } else { return true; } } return false; } 1. 2. 3. 4...
https://leetcode.cn/problems/subarrays-distinct-element-sum-of-squares-ii 如果求的是“和”而不是“平方和”,那就和周赛291最后一题一样,连Hard都达不到。不过“平方和”问题就复杂很多,不但需要lazy线段树这一高级数据结构来做区间更新,而且有一定的数学成分,需要在维护区间平方和的同时,保证区间和也能够快...
public int numSquares(int n) { // four-square and three-square theorems. while (n % 4 == 0) n /= 4; if (n % 8 == 7) return 4; if (this.isSquare(n)) return 1; // enumeration to check if the number can be decomposed into sum of two squares. ...
isSquare(n)) return 1; // enumeration to check if the number can be decomposed into sum of two squares. for (int i = 1; i * i <= n; ++i) { if (this.isSquare(n - i * i)) return 2; } // bottom case of three-square theorem. return 3; } } 复杂度分析...
167. Two Sum II - Input array is sorted 977. Squares of a Sorted Array (很像merge sort里的...
leetcode最经典100题 以下是LeetCode的最经典的100题：1.两数之和(Two Sum)2.两数相加(Add Two Numbers)3.无重复字符的最长子串(Longest Substring Without Repeating Characters)4.寻找两个有序数组的中位数(Median of Two Sorted Arrays)5.最长回文子串(Longest Palindromic Substring)6. Z字形变换(ZigZag ...
给定一个含有 n 个正整数的数组和一个正整数 target 。 找出该数组中满足其和 ≥ target 的长度最小的 连续子数组 [numsl, numsl+1, ..., numsr-1, numsr] ,并返回其长度。如果不存在符合条件的子数组,返回 0 。 代码语言:javascript 代码运行次数:0 ...
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