LeetCode 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

分析：设立cnt数组，表示在nums[i]处当前连续的1的值，maxn取其最大的值，在遇到nums[i] == 0的时候更新maxn的值。最后还要更新一下防止最后一个是1.

class Solution {
public:
    int findMaxConsecutiveOnes(vector<int>& nums) {
        vector<int> cnt(nums.size());
        int maxn = 0;
        cnt[0] = nums[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;
    }
};

class Solution {

public:

int findMaxConsecutiveOnes(vector<int>& nums) {

vector<int> cnt(nums.size());

int maxn = 0;

cnt[0] = nums[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;

}

};

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