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...
1、题目描述 2、问题分析 遍历一次数组,以每个1 为起点向后数,数到0 时比较当前1的个数和最大1 的个数,然后将遍历的起点放到当前0 的后面。 3、代码 1intfindMaxConsecutiveOnes(vector<int>&nums) {2intresult =0;3if( nums.size() ==0)returnresult;45for(inti =0; i < nums.size() ; i++)...
遍历数组进行判断取最大的连续1的个数即可。 classSolution {public:intfindMaxConsecutiveOnes(vector<int>&nums) {intres =0, cnt =0;for(inti =0; i != nums.size(); i++) {if(nums[i] ==1) cnt++;elsecnt=0; res=max(res, cnt); }returnres; } };//36 ms...
FindHeaderBarSize FindTabBarSize FindBorderBarSize Given a binary arraynums, returnthe maximum number of consecutive1's in the array. Example 1: Input:nums = [1,1,0,1,1,1]Output:3Explanation:The first two digits or the last three digits are consecutive 1s. The maximum number of consecuti...
LeetCode - 485. Max Consecutive Ones Q: Given a binary array, find the maximum number of consecutive 1s in this array. Example 1: Note: The input array will only contain 0 and 1. The length of input array is a positive integer and will n......
Leetcode 485: Max Consecutive Ones 问题描述: Given a binary array nums, return the maximum number of consecutive 1’s in the array. 最长连续的1 思路: 思路有好多种,可以利用数值特征用偏数学计算的方法,也可以用双指针。这里我采用前者: 遇到1就加和,遇到零就清空和,最大的和就是最长的1 代码:...
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....
题目地址:https://leetcode.com/problems/max-consecutive-ones-iii/ 题目描述 Given an arrayAof 0s and 1s, we may change up toKvalues from 0 to 1. Return the length of the longest (contiguous) subarray that contains only 1s. Example 1: ...
// #485 Description: Max Consecutive Ones | LeetCode OJ 解法1:水题。 // Solution 1: Trivial. 代码1 // Code 1 486 Predict the Winner // #486 预测赢家 描述:给定一个数组,每个元素表示一个分数。俩人轮流,每次从左端或右端拿走一个分数,如此直到拿完。分高的人赢。问先手是否必胜? // #486 ...
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...