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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.这一题看着简单,其实不是那么容易AC的。主要是需要对一些特殊的输入进行特殊处理。
注意:[2,1,1]这种情况是不存在第三大的,返回2。用三个变量first,second,third来表示第一大,第二大,第三大。
用count来表示这三个更新次数。如果更新次数大于等于3,直接返回third。 如果小于3,那就是返回最大值。 但是有一个特殊的例子: [2,1,-2147483648],这个在判断第三个时是不更新的,所以额外对其进行计数,用bottom来计数,在最后的判断中就用count+bottom来判断。 AC代码:9msclass Solution {public: int thirdMax(vector & nums) { #define INT_MIN (-2147483647-1) int max_3[3]; if(nums.size()==1) return nums[0]; else if(nums.size()==2) return nums[1]>nums[0]?nums[1]:nums[0]; else { int first=INT_MIN,second=INT_MIN,third=INT_MIN; bool flag1=0,flag2=0,flag3=0;//是否顶替掉初值 int count=0; int bottom=0; //缺少对三个一样的进行判断 for(int i=0;ifirst) { third=second; second=first; first=nums[i]; count++; } else if(nums[i]>second) { third=second; second=nums[i]; count++; } else if(nums[i]>third) { third=nums[i]; count++; } } if(count+bottom>=3)// return third;//三个都更新了 else return first;//两个更新,一个更新或者都没更新,没达到要求,返回第一个值 } }};
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