首先想到的是排序然后进行相应的查找，那么复杂度肯定就满足不了了
public static int thirdMax(int[] nums) {
if(nums.length == 0){
return 0;
}
if(nums.length == 1){
return nums[0];
}
if(nums.length == 2){
return Math.max(nums[0], nums[1]);
}
Arrays.sort(nums);
int count = 0;
for (int i = nums.length-1; i >= 0; i--) {
int j = nums[i];
if(i == 0){
}else {
if(nums[i] == nums[i-1]){
continue;
}else {
count ++;
if(count == 3){
return nums[i];
}
}
}
}
System.out.println(nums[nums.length-1]);
return nums[nums.length-1];
}
重新思考，这种算法，可能对于大多数情况是适用的，但是试想一下如果数组里面包含了最小值呢，可能就会报错。。。
class Solution {
public int thirdMax(int[] nums) {
int max1 = Integer.MIN_VALUE;
int max2 = Integer.MIN_VALUE;
int max3 = Integer.MIN_VALUE;
for (int i = 0; i < nums.length; i++) {
int j = nums[i];
if(max1 < j){
max3 = max2;
max2 = max1;
max1 = j;
}else if (j > max2 && j < max1) {
max3 = max2;
max2 = j;
}else if (j < max2&& j > max3) {
max3 = j;
}
}
if(max3 == Integer.MIN_VALUE){
return max1;
}else {
return max3;
}
}
}
转变一下思路，将最小值改成Long.MIN_VALUE即可
class Solution {
public int thirdMax(int[] nums) {
long max1 = Long.MIN_VALUE;
long max2 = Long.MIN_VALUE;
long max3 = Long.MIN_VALUE;
for (int i = 0; i < nums.length; i++) {
int j = nums[i];
if(max1 < j){
max3 = max2;
max2 = max1;
max1 = j;
}else if (j > max2 && j < max1) {
max3 = max2;
max2 = j;
}else if (j < max2&& j > max3) {
max3 = j;
}
}
if(max3 == Long.MIN_VALUE){
return (int) max1;
}else {
return (int) max3;
}
}
}