414. Third Maximum Number
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.
package leetcode.easy;
public class ThirdMaximumNumber {
@org.junit.Test
public void test() {
int[] nums1 = { 3, 2, 1 };
int[] nums2 = { 1, 2 };
int[] nums3 = { 2, 2, 3, 1 };
System.out.println(thirdMax(nums1));
System.out.println(thirdMax(nums2));
System.out.println(thirdMax(nums3));
}
public int thirdMax(int[] nums) {
int max1 = Integer.MIN_VALUE, max2 = Integer.MIN_VALUE, max3 = Integer.MIN_VALUE;
boolean minPresent = false;
for (int n : nums) {
if (n == Integer.MIN_VALUE) {
minPresent = true;
}
if (n > max1) {
max3 = max2;
max2 = max1;
max1 = n;
} else if (n > max2 && n < max1) {
max3 = max2;
max2 = n;
} else if (n > max3 && n < max2) {
max3 = n;
}
}
if (max3 != Integer.MIN_VALUE) {
return max3;
} else {
if (minPresent && max2 != Integer.MIN_VALUE) {
return max3;
} else {
return max1;
}
}
}
}
LeetCode_414. Third Maximum Number
原文：https://www.cnblogs.com/denggelin/p/11935413.html