leetcode-485. Max Consecutive Ones

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.
    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

题目很简单，就是求最大连续1的时候，用库会快很多。。第一种方法太傻了o(╥﹏╥)o

class Solution {
    public int findMaxConsecutiveOnes(int[] nums) {
                    int[] solu = new int[nums.length];
            int count = 0,flag = 0;
            for (int i = 0; i < nums.length; i++) {
                if(nums[i]==1){
                    for (int j = i;j<nums.length;j++){
                        if(nums[j]==1&&j != nums.length -1) {
                            count++;                          
                        }
                        else if(nums[j] == 1&& j == nums.length -1){
                            count++;
                            solu[flag] = count;
                            flag++;
                        }
                        else{
                            //list.add(count);
                            solu[flag] = count;
                            flag++;
                            break;
                        }

                    }
                    count = 0;
                }
            }
            Arrays.sort(solu);
            return solu[solu.length-1];
    }
}

Runtime: 510 ms

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

Runtime: 11 ms