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...
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......
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
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python Solution LeetCode No.485 Max Consecutive Ones (最大连续1的个数) 题目: 给定一个二进制数组, 计算其中最大连续1的个数。 示例 1: 输入: [1,1,0,1,1,1] 输出: 3 解释: 开头的两位和最后的三位都是连续1,所以最大连续1的个数是 3. 注意: 输入的数组只包含 0 和1。 输入数组的长度...
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: ...
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 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 ...