题目地址:https://leetcode-cn.com/problems/max-consecutive-ones-ii/题目描述Given a binary array, find the maximum number of consecutive 1s in this array if you can flip at most one 0.Example 1:Input: [1,0,1,1,0] Output: 4 Explanation: Flip the first zero will get the the maximum ...
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....
Can you solve this real interview question? Max Consecutive Ones - Given a binary array nums, return the maximum number of consecutive 1's in the array. Example 1: Input: nums = [1,1,0,1,1,1] Output: 3 Explanation: The first two digits or the last
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...
leetcode 485. Max Consecutive Ones Given a binary array, find the maximum number of consecutive 1s in this array. Example 1: Input: [1,1,0,1,1,1] Output: 3 Explanation: The first two digits or the last three digits are consecutive 1s....
Max Points on a Line 11.2% Hard LRU Cache 14.1% Hard Longest Valid Parentheses 19.7% Hard Longest Consecutive Sequence 28.2% Hard Copy List with Random Pointer 23.5% Hard Largest Rectangle in Histogram 21.5% Hard Jump Game II 24.7% Hard Interleaving String 19.5% Hard Insert Interval 20.7% Hard...
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...
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...
1. 根据问题的不同,将 K 个元素插入到 min-heap 或 max-heap 中 2.迭代处理剩余的数,如果你找到一个比 heap 中数更大的数,那么就移除那个数并插入这个更大的数 这里无需排序算法,因为 heap 将为你跟踪这些元素。 如何识别前 K 个元素模式: 如果你被要求寻找一个给定集合中前面的/最小的/最常出现的 ...
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...