leetcode题解-485. Max Consecutive Ones

题目：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

本题目的是找出一个二进制数组中最长的连续1的个数。我们可以使用两个指针来标识当前连续的1的子串，用max记录目前为止最大的连续长度。代码入下：

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

此外，因为题目说了是01数组，所以我们还可以使用乘和加的方式。每当遇到0的时候，sum就会变成0，如果全是1，sum就会一直叠加。代码如下所示：

    public int findMaxConsecutiveOnes1(int[] nums) {        int maxSum = 0, sum = 0;        for (int n : nums) {            sum *= n;            sum += n;            maxSum = Math.max(maxSum, sum);        }        return maxSum;    }