300. Longest Increasing Subsequence

Given an unsorted array of integers, find the length of longest increasing subsequence.

For example,
Given [10, 9, 2, 5, 3, 7, 101, 18],
The longest increasing subsequence is [2, 3, 7, 101], therefore the length is 4. Note that there may be more than one LIS combination, it is only necessary for you to return the length.

Your algorithm should run in O(n2) complexity.

Follow up: Could you improve it to O(n log n) time complexity?

题目含义：给定一个没有排序的数组，返回最长递增子序列（注意不是子字符串）的长度

    public int lengthOfLIS(int[] nums) {
        if(nums.length == 0){
            return 0;
        }
        int[] a = new int[nums.length];
        int max = 0;
        //依次遍历每一个元素，如果前面没有比他小的数字，则该数字构成的子串最大长度为1.
        // 如果前面有多个比他小的数字，找出他们的最大值，然后加1作为本节点的最大长度
        for (int i=0;i<nums.length;i++)
        {
            a[i] = 1;
            for (int j=0;j<i;j++)
            {
                if(nums[j]<nums[i])
                {
                    a[i] = Math.max(a[i],a[j]+1);
                }
            }
            max = Math.max(max,a[i]);
        }
        return max;        
    }