Howard的程式筆記

485. Max Consecutive Ones (連續出現最多次數的1)

分類於

題目

Given a binary array nums, return the maximum number of consecutive 1’s in the array.

返回陣列中連續出現最多次數的1

Example 1

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Input: nums = [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.

Example 2

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Input: nums = [1,0,1,1,0,1]
Output: 2

Constraints:

  • 1 <= nums.length <= 105
  • nums[i] is either 0 or 1.

我的解題

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class Solution {
    public int findMaxConsecutiveOnes(int[] nums) {
        int tmpNum = 0;		//暫存數
        int maxNum = 0;		//最大數
        for(int i=0;i<nums.length;i++){
            if(nums[i] == 1){
                tmpNum++;		//若nums[i]的值==1，則暫存數+1
                maxNum = Math.max(tmpNum,maxNum);		//比較兩值並將較大的存進"最大數"
			}else{
                tmpNum = 0;		//若nums[i]的值!=1，則暫存數=0
			}
        }
        return maxNum;		//回傳最大數
    }
}