标签：== HERE strong uniq NPU The inpu ide rem
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.
Idea. Unique is the key, at first tried Integer.MIN_VALUE to initialise the array, more work to tell if the max3 is MIN_VALUE or not, use null instead
Time complexity: O(N)
Space complexity: O(1)
1 class Solution { 2 public int thirdMax(int[] nums) { 3 Integer max1 = null; 4 Integer max2 = null; 5 Integer max3 = null; 6 7 for(int num: nums) { 8 if(max1 == null || num > max1) { 9 max3 = max2; 10 max2 = max1; 11 max1 = num; 12 } 13 else if(num != max1 && (max2 == null || num > max2)) { 14 max3 = max2; 15 max2 = num; 16 } 17 else if((num != max1 && num != max2) && (max3 == null || (num > max3))) { 18 max3 = num; 19 } 20 } 21 22 if(max3 == null) { 23 return max1; 24 } 25 return max3; 26 } 27 }
Use Integer to wrap the array, to remove the duplicates in max1, max2 and max3, avoid the long if conidition
1 class Solution { 2 public int thirdMax(int[] nums) { 3 Integer max1 = null; 4 Integer max2 = null; 5 Integer max3 = null; 6 7 for(Integer num: nums) { 8 if(num.equals(max1) || num.equals(max2) || num.equals(max3)) { 9 continue; 10 } 11 12 if(max1 == null || num > max1) { 13 max3 = max2; 14 max2 = max1; 15 max1 = num; 16 } 17 else if(max2 == null || num > max2) { 18 max3 = max2; 19 max2 = num; 20 } 21 else if(max3 == null || num > max3) { 22 max3 = num; 23 } 24 } 25 26 if(max3 == null) { 27 return max1; 28 } 29 return max3; 30 } 31 }
Another idea, use Long.MIN_VALUE instead
1 class Solution { 2 public int thirdMax(int[] nums) { 3 long max1 = Long.MIN_VALUE; 4 long max2 = Long.MIN_VALUE; 5 long max3 = Long.MIN_VALUE; 6 7 for(int num: nums) { 8 if(num > max1) { 9 max3 = max2; 10 max2 = max1; 11 max1 = num; 12 } 13 else if(num != max1 && num > max2) { 14 max3 = max2; 15 max2 = num; 16 } 17 else if((num != max1 && num != max2) && num > max3) { 18 max3 = num; 19 } 20 } 21 22 return max3 == Long.MIN_VALUE? (int)max1 : (int)max3; 23 } 24 }
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.
标签：== HERE strong uniq NPU The inpu ide rem
原文地址：https://www.cnblogs.com/taste-it-own-it-love-it/p/10850680.html
