3 LeetCode 485 3.1 题目 先上英文版的题目,再看中文版。 3.2 解法一:逐个元素判断 fromtypingimportListclassSolution:deffindMaxConsecutiveOnes(self,nums:List[int])->int:##注意这里是“List“,而不是”list“;ifnumsisNoneorlen(nums)==0:## 如果列表 lis 为空return0count=result=0## 计数变量,结果...
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++){6if( nums...
(leetcode题解)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. The maximum number of consecutive 1s is 3. Not...
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...
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 - 128. Longest Consecutive Sequence (哈希表) LeetCode - 128. Longest Consecutive Sequence (哈希表) 题目链接 题目 解析 第一种方法: 使用一个HashSet来存储对应的值,一开始先将所有的值都加入set; 遍历数组的每一个元素,每次去检查当前元素num的前一个元素num - 1是不是在set中,如果是,说明...
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[LeetCode]485. Max Consecutive Ones 找到最大的连续的1的个数 题目描述 输入只有0和1的数组(长度为正整数,且<10000),找到最大的连续1的个数 比如[1,1,0,1,1,1],输出3 思路 遍历数组,统计当前连续个数curCount和最大连续值maxCount。If当前数组值为1,则curCount++;否则(值为0)比较curCount和max...
LeetCode之Max Consecutive Ones 1、题目 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 Consecutive Ones II 题目链接:https://leetcode.com/problems... public class Solution { public int findMaxConsecutiveOnes(int[] nums) { // 2 points, slide window int i = 0, j = 0; int global = 0; // count the number of flip ...