1、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
这个题直观看上去的思路就是，从头到尾遍历，设一个变量存最大值，设一个临时变量设当前值，对比当前和最大，交换or not，遇到0就更新临时变量为0就好了。
class Solution {public: int findMaxConsecutiveOnes(vector<int>& nums) { int max=0; int s=nums.size(); int count=0; for(int i=0;i<s;i++){ if(nums[i]==1){ count++; } else if(nums[i]==0&&count!=0){ count=0; } if(count>max) {max=count;} } return max; }};不过仔细一想，数组的数比较特殊，只有0和1，可借助乘0可以达到归零的效果，＋1可以达到计数的作用。
class Solution {public: int findMaxConsecutiveOnes(vector<int> nums) { int max = 0; int sum = 0; for (int i=0; i<nums.size(); i++) { sum = (sum+nums[i])*nums[i]; if(max<sum){max=sum;} } return max; }};新闻热点
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