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:
0 and 1.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
原文：http://www.cnblogs.com/qinduanyinghua/p/6350073.html