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【42】414. Third Maximum Number
414. Third Maximum Number
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- Total Accepted: 20624
- Total Submissions: 76932
- Difficulty: Easy
- Contributors: ZengRed , 1337c0d3r
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.
1 class Solution { 2 public: 3 int thirdMax(vector<int>& nums) { 4 long first = LONG_MIN, second = LONG_MIN, third = LONG_MIN;//初始化要用长整型long的最小值，否则当数组中有INT_MIN存在时，程序就不知道该返回INT_MIN还是最大值first了 5 for(int i = 0; i < nums.size(); i++){ 6 if(nums[i] > first){//update 3 个 7 third = second; 8 second = first; 9 first = nums[i]; 10 }else if(nums[i] > second && nums[i] < first){//update second & third 11 third = second; 12 second = nums[i]; 13 }else if(nums[i] > third && nums[i] < second){ 14 third = nums[i]; 15 } 16 } 17 /* 18 if(third == INT_MIN){ 19 return second == INT_MIN ? first : second; 20 } 21 return third;*/ 22 return (third == LONG_MIN || third == second) ? first : third; 23 } 24 };
【42】414. Third Maximum Number
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