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;
}
}