Problem:
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. Code: public class Solution { public int findMaxConsecutiveOnes(int[] nums) { int count = 0; int count2 = 0; for(int i = 0; i < nums.length; i++){ if (nums[i] == 0){ count2 = 0; } if (nums[i] == 1){ count2 += 1; } count = Math.max(count,count2); } return count; } }
Code2:
public class Solution { public int findMaxConsecutiveOnes(int[] nums) { int count = 0; int count2 = 0; for(int n : nums){ count = Math.max(count,count2 = n == 0 ? 0 : count2 + 1); //高手倾向于用增强型for循环和三目运算符写 } return count; } }