題目

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 number of consecutive 1s is 3.

Note:

• The input array will only contain 0 and 1.
• The length of input array is a positive integer and will not exceed 10,000

分析

二進制數組中最多的連續'1'的個數

解答

解法1：（我）每次'0'時取max，但返回需要再取一次max（12ms）

public class Solution {public int findMaxConsecutiveOnes(int[] nums) {int count = 0;int count1 = 0;for (int i = 0; i < nums.length; i++){if(nums[i] == 1){count++;}else{count1 = Math.max(count,count1);count = 0;}}return Math.max(count,count1);} }

?

解法2：每次'1'時取max，返回無需再取；遍歷數組由for改為for each（9ms√）

public class Solution {public int findMaxConsecutiveOnes(int[] nums) {int count = 0;int result = 0;for (int num : nums){if(num == 1){count++;result = Math.max(count,result);}else{count = 0;}}return result;} }

轉載于:https://www.cnblogs.com/xuehaoyue/p/6412584.html

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