LeetCode | 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

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老老實實按位算就行，性能較低，需要判斷的條件較多，only beats 14%

public class Solution {
    public int findMaxConsecutiveOnes(int[] nums) {
        int max = 0;
        int count = 0;
        for(int i = 0; i < nums.length; i++) {
            if(nums[i] == 1) {
                count++;
                if(i == nums.length - 1) {
                    max = count > max? count : max;
                }
            } else {
                max = count > max? count : max;
                count = 0;
            }
        }
        return max;
    }
}

異曲同工的另一種解法，只是代碼更加簡潔

public class Solution {
    public int findMaxConsecutiveOnes(int[] nums) {
        int max = 0;
        int count = 0;
        for(int n : nums)
            max = Math.max(max, count = n == 0?0 : ++count);
        return max;
    }
}

下面提供另一種通過 String API 來解決的方法，就是將數組轉換爲字符串，再利用split方法轉換爲多個只含1序列的字符串，找出最長的即答案，不過LeetCode一直報超時，測試卻沒問題。曬出代碼僅供參考。

public class Solution {
    public int findMaxConsecutiveOnes(int[] nums) {
        String str = "";
        int max = 0;
        for(int n : nums) str = str + n;
        String[] ones = str.split("0");
        for(String tmp : ones) max = Math.max(max, tmp.length());
        return max;
    }
}