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

Input: [10,9,2,5,3,7,101,18]
Output: 4 
Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4. 

Note:
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?

难度：medium

题目：给定一无序的整数数组，找出其递增的最大子序列。

思路：动态规划， 用一个长度与输入数组相同长度的数组记录从0到当前元素的最大递增元素个数。
dp(i) = Math.max(dp[j]…) + 1, (nums[i] > nums[j])
dp(i) = 1 (nums[i] <= nums[j], 0 <= j <= i – 1)

class Solution {
    public int lengthOfLIS(int[] nums) {
        int n = nums.length, maxIncLength = 0;
        int[] incLength = new int[n];
        for (int i = 0; i < n; i++) {
            int curMaxIncLength = 0;
            for (int j = 0; j <= i; j++) {
                if (nums[j] < nums[i]) {
                    curMaxIncLength = Math.max(curMaxIncLength, incLength[j]);
                }
            }
            incLength[i] = curMaxIncLength + 1;
            maxIncLength = Math.max(maxIncLength, incLength[i]);
        }
     
        return maxIncLength;
    }
}