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.
这道题的test case写的很巧妙
public class Solution {
public int thirdMax(int[] nums) {
Integer f = null;
Integer s = null;
Integer result = null;
for(int i=0;i<nums.length;i++){
System.out.println(nums[i]);
if((f != null && nums[i] ==f) || (s != null && nums[i] ==s) || (result != null && nums[i] ==result)){
continue;
}
if(f == null || nums[i]>f){
result = s;
s = f;
f = nums[i];
}
else if(s == null || nums[i]>s){
result = s;
s = nums[i];
}
else if(result == null || nums[i]>result){
result = nums[i];
}
}
if(nums.length <3 || result == null){
return f;
}
return result;
}
}
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