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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

题意：找出数组中连续为1的长度n；

思路：

　　　1.遍历数组，创建一个计数器和一个结果result，如果数组元素等于1，则count++；求出result和count的最大值赋值为result，否则count=0；重新开始计数；

　　　2.巧妙利用0乘于任何数等于0的事实；

1.代码实现：

class Solution {
    public int findMaxConsecutiveOnes(int[] nums) {
        int count = 0;
        int result = 0;
        int len = nums.length;
        for(int i = 0;i < len; i++){
            if(nums[i]==1){
                count++;        //计数器
                result = Math.max(count,result);    result和count的最大值
            }else{
                count = 0;
            }
        }
        return result;
    }
}

2.代码实现：

int findMaxConsecutiveOnes(int* nums, int numsSize) {
    int i;
    int sum = 0,max = 0;
    for(i = 0;i<numsSize;i++){
        sum =(sum + nums[i])*nums[i];    //求出sum值，sum相当于计数器
        if(max<sum){
            max = sum;
        }
    }
    return max;
}