[抄题]:
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.
[暴力解法]:
用temp,result,结果发现写起来很麻烦
时间分析:
空间分析:
[优化后]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
[思维问题]:
[一句话思路]:
还是局部最大值+全局最大值的思路。用三元表达式解决局部变量的连续和中断问题
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
[一刷]:
- 是maxHere + 1 统计,不是n + 1,这是是乐至?
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
用三元表达式解决局部变量的连续和中断问题
[复杂度]:Time complexity: O(n) Space complexity: O(1)
[英文数据结构或算法,为什么不用别的数据结构或算法]:
[关键模板化代码]:
[其他解法]:
[Follow Up]:
487. Max Consecutive Ones II 好像不是数学就是两根指针吧
[LC给出的题目变变变]:
[代码风格] :
class Solution { public int findMaxConsecutiveOnes(int[] nums) { //cc if (nums == null || nums.length == 0) { return 0; } //ini int max = 0; int maxHere = 0; //for loop for (int n : nums) { max = Math.max(max, maxHere = n == 0 ? 0 : maxHere + 1); } //return return max; } }
原文地址:https://www.cnblogs.com/immiao0319/p/8893855.html
时间: 2024-11-08 22:33:30