【LeetCode】数组-9(414)-O(n)内找到第三大的数

思路：

　　从前向后遍历，用三个变量first second third 保存前三个大的数，初值设为long类型的无穷小（因为开始提交到案遇到负的临界值的情况），如果新来的数大于first则 second first依次后移并且把这歌新值赋值给first

【正确代码】

class Solution {
    public int thirdMax(int[] nums) {
        long first = nums[0], second = Long.MIN_VALUE, third = Long.MIN_VALUE;
        if (nums.length == 1) {
            return (int)nums[0];
        }
        if (nums.length == 2) {
            return nums[0] > nums[1] ? (int)nums[0] : (int)nums[1];
        }
        int count = 0;
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] > first) {
                third = second;
                second = first;
                first = nums[i];
            }else if (nums[i] > second && nums[i] < first) {
                third = second;
                second = nums[i];
            }else if (nums[i] < second && nums[i] > third) {
                third = nums[i];
            }else {
                continue;
            }
            count++;
        }
        if (count < 2) {
            return (int)first;
        }
        return (int)third;
    }
}

有时间想想用treeset的容器做一下。