LeetCode: 485 Max Consecutive Ones(easy)

原创于 2019-04-18 10:20:22 发布 · 262 阅读

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本文介绍了一种在二进制数组中查找最长连续1序列的方法。通过遍历数组，使用计数器跟踪当前连续1的数量，并更新最大连续1的记录。此算法适用于仅包含0和1的输入数组，长度为正整数且不超过10,000。

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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

代码：

别人的：

class Solution {
public:
    int findMaxConsecutiveOnes(vector<int>& nums) {
        int count = 0, max = 0;
        for (int i = 0; i < nums.size(); i++) {
            if (nums[i] == 1 && (!i || nums[i - 1] != 1)) count = 1;
            else if (i && nums[i] == 1 && nums[i - 1] == 1) count++;
            if (max < count) max = count;
        }
        if (max < count) max = count;
        return max;
    }
};

自己的：

class Solution {
public:
    int findMaxConsecutiveOnes(vector<int>& nums) {
        int result = 0;
        int tem = 0;
        nums.push_back(0);
        for (auto c : nums){
            if(c == 0){
                if ( tem > result)
                    result = tem;
                tem = 0;
            }
            else
                tem++;
        }
        return result;
    }
};