[抄题]:
Given a non-empty array of integers, return the third maximum number in this array. If it does not exist, return the maximum number. The time complexity must be in O(n).
Example 1:
Input: [3, 2, 1] Output: 1 Explanation: The third maximum is 1.
Example 2:
Input: [1, 2] Output: 2 Explanation: The third maximum does not exist, so the maximum (2) is returned instead.
Example 3:
Input: [2, 2, 3, 1] Output: 1 Explanation: Note that the third maximum here means the third maximum distinct number. Both numbers with value 2 are both considered as second maximum.
[暴力解法]:
时间分析:nlgn
空间分析:
[优化后]:
时间分析:n
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
[思维问题]:
[一句话思路]:
不允许排序,就只能一个个地放了
[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]:
用else if :
[一刷]:
- 去重复的方法:用continue继续处理,好像很少用。
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
同时判断要用else if
[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]:
- 不能初始化为0,就包装一下,初始化为null
- 或者初始化为极值
[关键模板化代码]:
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
class Solution { public int thirdMax(int[] nums) { //ini Integer max1 = null; Integer max2 = null; Integer max3 = null; //for loop, change for (Integer n : nums) { if (n.equals(max1) || n.equals(max2) || n.equals(max3)) continue; if (max1 == null || n > max1) { max3 = max2; max2 = max1; max1 = n; } else if (max2 == null || n > max2) { max3 = max2; max2 = n; } else if (max3 == null || n > max3) { max3 = n; } } //return return (max3 == null) ? max1 : max3; } }
[代码风格] :