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.
我的答案:
class Solution {
public int thirdMax(int[] nums) {
sort(nums);
int a = 1;
int thrid = nums[nums.length - 1];
for (int i = nums.length-1; i >=0; i--) {
int num = nums[i];
if (num < thrid) {
a++;
thrid = num;
}
if (a == 3) {
break;
}
}
if (a < 3) {
return nums[nums.length - 1];
}
return thrid;
}
private static int partition(int[] arr, int l, int r){
int v = arr[l];
int j = l; // arr[l+1...j] < v ; arr[j+1...i) > v
for( int i = l + 1 ; i <= r ; i ++ )
if( arr[i]< v ){
j ++;
swap(arr, j, i);
}
swap(arr, l, j);
return j;
}
private static void sort(int[] arr, int l, int r){
if( l >= r )
return;
int p = partition(arr, l, r);
sort(arr, l, p-1 );
sort(arr, p+1, r);
}
public static void sort(int[] arr){
int n = arr.length;
sort(arr, 0, n-1);
}
private static void swap(int[] arr, int i, int j) {
int t = arr[i];
arr[i] = arr[j];
arr[j] = t;
}
}
大神的答案:
public int thirdMax(int[] nums) {
Integer max1 = null;
Integer max2 = null;
Integer max3 = null;
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 max3 == null ? max1 : max3;
}