[Leetcode] 485. Max Consecutive Ones 解题报告

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: 3Explanation: 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

思路：

一道练手题目。不过看到网上写的代码真的很精妙，所以再贴一下。但是两个代码的时间复杂度都是O(n)，空间复杂度都是O(1)。

代码：

1、一般解法：

class Solution {public:    int findMaxConsecutiveOnes(vector<int>& nums) {        int max_num = 0, num = 0;        bool started = false;        for (int i = 0; i < nums.size(); ++i) {            if (nums[i] == 1) {                if (!started) {                    started = true;                    num = 1;                }                else {                    ++num;                }            }            else {                if (started) {                    max_num = max(max_num, num);                    started = false;                    num = 0;                }            }        }        return max(max_num, num);    }};

2、精妙解法：

class Solution {public:    int findMaxConsecutiveOnes(vector<int>& nums) {        int ret = 0, sum = 0;        for (int i = 0; i < nums.size(); ++i) {            sum = (sum + nums[i] ) * nums[i];            ret = max(ret, sum);        }        return ret;    }};