如果当前数字为奇数,则将其加上 1 。 题目保证你总是可以按上述规则将测试用例变为 1 。 来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/number-of-steps-to-reduce-a-number-in-binary-representation-to-one 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。 思路 既然处...
**解析:**Version 1,这道题跟Leetcode 560的解法很像,首先计算数组的总和total,如果total < x,则无论如何也不会将x减到0,如果total = x,则需要移除所有元素才能将x变为0,由于x一直是从最左或最右移除,因此问题可以变为:找到一个最大连续子数组,使得其和为total - x,这样可以保证剩下的元素之和等于x,...
Return the minimum size of the set so that at least half of the integers of the array are removed. Example 1: Input: arr = [3,3,3,3,5,5,5,2,2,7] Output: 2 Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the...
题内话,大模型相关的岗位,手撕CUDA的概率非常大,leetcode反而写的少,就前段时间个人的经验,基本是4:1的比例,还是建议好好复习下CUDA。当然,这些只是最简单的kernel实现,比如flash_attn,FMHA, FMHCA这些优化手段,就不在这篇文章里写了,面试中基本都会问到。后边有空再补档一些文章吧。
Explanation: The optimal solution is to remove the last three elements and the first two elements (5 operations in total) to reduce x to zero. Constraints: 1 <= nums.length <= 105 1 <= nums[i] <= 104 1 <= x <= 109 从给定数组左边和右边取连续的数,问最少的取数个数能让和等于x...
1. Description Reduce Array Size to The Half 2. Solution 解析:对数组中的元素个数进行统计,并按从大到小排列,一次移除一个元素,直至剩余元素个数小于等于移除的元素个数。 Version 1 classSolution:defminSetSize(self,arr:List[int])->int:stat={}n=len(arr)fornuminarr:stat[num]=stat.get(num,0)...
1. Description Minimum Operations to Reduce X to Zero 2. Solution 解析:Version 1,这道题跟Leetcode 560的解法很像,首先计算数组的总和total,如果total < x,则无论如何也不会将x减到0,如果total = x,则需要移除所有元素才能将x变为0,由于x一直是从最左或最右移除,因此问题可以变为:找到一个最大连续子...
classSolution{public:intminOperations(vector<int>& nums,intx){intmin_step = INT_MAX;intsum = accumulate(nums.begin(), nums.end(),0);if(x > sum)return-1;intl =0, r =0, max_len =-1, cur_sum =0, n = nums.size();while(l < n){if(r < n) cur_sum += nums[r++];while...
package LeetCode_1342 /** * 1342. Number of Steps to Reduce a Number to Zero * https://leetcode.com/problems/number-of-steps-to-reduce-a-number-to-zero/ * Given a non-negative integer num, return the number of steps to reduce it to zero. * If the current number is even, you ...
fmt.Fprintf(os.Stderr,"Usage: mrsequential xxx.so inputfiles...\n") os.Exit(1) } mapf, reducef := loadPlugin(os.Args[1]) // // read each input file, // pass it to Map, // accumulate the intermediate Map output. //