classSolution {public:intfindMaxConsecutiveOnes(vector<int>&nums) {intres =0, zero =0, left =0, k =1;for(intright =0; right < nums.size(); ++right) {if(nums[right] ==0) ++zero;while(zero >k) {if(nums[left++] ==0) --zero; } res= max(res, right - left +1); }return...
1classSolution {2publicintfindMaxConsecutiveOnes(int[] nums) {3intmax = 0;4//保持queue中只有一个15//等价于滑动窗口找子串,子串内最多只能有一个16intk = 1;7Queue<Integer> queue =newLinkedList<>();8for(intl = 0, h = 0; h < nums.length; h++) {9if(nums[h] == 0) {10queue....
然后用result的存储最大的count,并和最后的count对比,返回最大的count## 精简版本classSolution:deffindMaxConsecutiveOnes(self,nums:List[int])->int:max_count=count=0fornuminnums:ifnum==1:count+=1max_count=max(max_count,count)else:count=0# 无视0即可returnmax_count...
class Solution { public int findMaxConsecutiveOnes(int[] nums) { int max=0,curr=0; for(int iterator:nums){ if(iterator==0){ curr=0; }else{ curr++; if(curr>max) max=curr; } } return max; } } 这个方法只需遍历一遍数组,accept之后显示runtime为9ms 更好的办法 提交了之后发现一个run...
// #487 Description: Max Consecutive Ones II | LeetCode OJ 解法1:搞个状态机试试? // Solution 1: Try using a state machine? 代码1 // Code 1 488 Zuma Game // #488 祖玛游戏 描述:给定一串彩色球,再给你几个球用来发射。每次发射你都可以随便挑剩下的任一个,3个一消。问最少用多少个球可...
iterate nums, 是1, curMax++, 否则curMax 清0, 时刻维护最大值. Time Complexity: O(nums.length). Space: O(1). AC Java: AI检测代码解析 1publicclassSolution {2publicintfindMaxConsecutiveOnes(int[] nums) {3intres = 0;4intcurMax = 0;5for(intnum : nums){6curMax = num == 1 ? cur...
485. Max Consecutive Ones solution1: AI检测代码解析 class Solution { public: int findMaxConsecutiveOnes(vector<int>& nums) { if(nums.empty()) return 0; int ans = 0, max = INT_MIN; for(int i=0; i<nums.size(); i++) ...
C++ 智能模式 1 2 3 4 5 6 class Solution { public: int findMaxConsecutiveOnes(vector<int>& nums) { } }; 已存储 行1,列 1 运行和提交代码需要登录 Case 1Case 2 nums = [1,1,0,1,1,1] 1 2 [1,1,0,1,1,1] [1,0,1,1,0,1] Source ...
487.Max-Consecutive-Ones-II (H-) 1186.Maximum-Subarray-Sum-with-One-Deletion (H-) 1187.Make-Array-Strictly-Increasing (H-) 1909.Remove-One-Element-to-Make-the-Array-Strictly-Increasing (H-) 3196.Maximize-Total-Cost-of-Alternating-Subarrays (M) 区间型 I 132.Palindrome-Partitioning-II (H...
487 Max Consecutive Ones II $ 42.70% Medium 486 Predict the Winner 43.60% Medium 485 Max Consecutive Ones 55.30% Easy 484 Find Permutation $ 50.50% Medium 483 Smallest Good Base 30.60% Hard 482 License Key Formatting 41.20% Medium 481 Magical String 46.20% Medium 480 Sliding Window Median 31.0...