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...
1publicclassSolution {2publicintfindMaxConsecutiveOnes(int[] nums) {3int[] newNums =newint[nums.length+2];4newNums[0] = 0;5newNums[newNums.length-1] = 0;6for(inti = 1; i < newNums.length-1; i++) {7newNums[i] = nums[i-1];8}9intans = 0;10intlastPos = 0;11for(inti...
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++)...
class Solution(object): def findMaxConsecutiveOnes(self, nums): """ :type nums: List[int] :rtype: int """ ans = 0 i = 0 length = len(nums) while i<length: ones = 0 while i<length and nums[i] == 1: i += 1 ones += 1 i += 1 # i == length or nums[i]==0 ans ...
The maximum number of consecutive 1s is 3. 1. 2. 3. 4. Note: The input array will only contain0and1. The length of input array is a positive integer and will not exceed 10,000 2、代码实现 public class Solution { public int findMaxConsecutiveOnes(int[] nums) { ...
def findMaxConsecutiveOnes(self, nums): """ :type nums: List[int] :rtype: int """ return len(max("".join(map(str, nums)).split("0"))) 别人解法思路: 设置两个计数器:结果值和计数值。以此访问list,如果值为1,计数值加一,结果值取结果值和计数值中的大值。如果值为0,计数值复为0。
[LeetCode] Max Consecutive Ones 最大连续1的个数 Given a binary array, find the maximum number of consecutive 1s in this array. Example 1: Input:[1,1,0,1,1,1]Output:3Explanation:The first two digits or the last three digits are consecutive 1s....
485. Max Consecutive Ones # 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. The maximum numbe
func findMaxConsecutiveOnes(_ nums: [Int]) -> Int { var count = 0,maxCount = 0 var i = 0 while i < nums.count { while i < nums.count && nums[i] == 1{ count += 1 i += 1 } if count > maxCount { maxCount = count ...
描述:给定一个单词,允许把其中一个或多个部分替换成对应长度,但替换的部分不能相邻。先给定一个单词和一个对应缩写,判断缩写是否合法。 //#408Description: Valid Word Abbreviation | LeetCode OJ 解法1:水题。 // Solution 1: Trivial. 代码1 //Code 1 ...