485. 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

c代码：

int findMaxConsecutiveOnes(int* nums, int numsSize) {
    int count = 0, max = 0;
    for(int i = 0; i < numsSize; i++){
        if(nums[i] == 1){
            count++;
        }
        else{
            if(count > max)
                max = count;
            count = 0;
        }
    }
    //如果最后一个是1，这个判断是很有必要的
    if(count > max)
        max = count;
    return max;
    
}

c++：

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

用向量做法：

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

没有必要用c++的，用c，通过运行时间，显然c快