但是Hackerrank调试器在解释这个错误时非常糟糕，所以我怀疑我的理解。无论如何，这是另一种不使用向量的...
classSolution {public:intminMoves2(vector<int>&nums) {intn =nums.size();if(n ==1)return0; sort(nums.begin(), nums.end());intmid = nums[n/2];intret =0;for(auto v : nums) ret += abs(v -mid);returnret; } };
One pass in-place solution: all swaps. classSolution {public:/** * @param nums: a vector of integers * @return: nothing*/voidpartitionArray(vector<int> &nums) { size_t len=nums.size();inti =0, io =0, ie = len -1;while(io <ie) {intv =nums[i];if(v &0x1)//odd{ swap(...
all(nums[i] >= nums[i +1] for i inrange(len(nums) -1))) print(solution(A)) print(solution(B)) print(solution(C)) Output: True False True 这也是一个常见的问题,此处提供的算法十分优雅,只用一行就可以写完。一个数组是单调的,当且仅当它单调增加或单调减少。为了判断数组是否单调,这个算法利...
2025年5月> 日一二三四五六 27282930123 45678910 11121314151617 18192021222324 25262728293031 1234567 Increasing all elements by 1 except one element, equals to decreasing that one element. classSolution {public:intminMoves(vector<int>&nums) {intminv = *min_element(nums.begin(), nums.end());intret...