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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: 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
0and1. - The length of input array is a positive integer and will not exceed 10,000
c代码：
1 int findMaxConsecutiveOnes(int* nums, int numsSize) { 2 int count = 0, max = 0; 3 for(int i = 0; i < numsSize; i++){ 4 if(nums[i] == 1){ 5 count++; 6 } 7 else{ 8 if(count > max) 9 max = count;10 count = 0;11 }12 }13 //如果最后一个是1，这个判断是很有必要的14 if(count > max)15 max = count;16 return max;17 18 }
c++：
1 class Solution { 2 public: 3 int findMaxConsecutiveOnes(vector<int>& nums) { 4 int count = 0, maxn = 0; 5 int size = nums.size(); 6 for(int i = 0; i < size; i++){ 7 if(nums[i] == 1){ 8 count++; 9 }10 else{11 maxn = max(count, maxn);12 count = 0;13 }14 }15 maxn = max(count, maxn);16 return maxn;17 }18 };
用向量做法：
1 class Solution { 2 public: 3 int findMaxConsecutiveOnes(vector<int>& nums) { 4 vector<int> cnt(nums.size()); 5 cnt[0] = nums[0]; 6 int maxn = 0; 7 for(int i = 1; i < nums.size(); i++){ 8 if(nums[i] == 0){ 9 cnt[i] = 0;10 maxn = max(maxn, cnt[i-1]);11 } else {12 cnt[i] = cnt[i-1] + 1;13 }14 }15 maxn = max(maxn, cnt[nums.size() - 1]);16 return maxn;17 }18 };
没有必要用c++的，用c，通过运行时间，显然c快
485. Max Consecutive Ones
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