Longest Increasing Subsequence

题目描述：

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?

这个题想就是用动态规划的方法去做，但是一开始我陷入了误区，我思考的是dp[i]要么等于dp[i-1]，要么等于dp[i-1]+1.这样讨论起来就比较复杂了。

这个题的递归式是：dp[i] = max{dp[j] + 1,dp[i]}     其中j < i && nums[j] < nums[i]

然后用自下而上的动态规划算法即可求解：

public class Solution {
    public int lengthOfLIS(int[] nums) {
        if(nums.length<2)
        	return nums.length;
        int[] dp=new int[nums.length];
        Arrays.fill(dp, 1);int max=1;
        for(int i=1;i<nums.length;i++){
        	for(int j=0;j<i;j++){
        		if(nums[j]<nums[i]&&dp[j]+1>dp[i]){
        			dp[i]=dp[j]+1;
        			if(dp[i]>max)
        				max=dp[i];
        		}
        	}
        }
        return max;
    }
}