300. Longest Increasing Subsequence

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 is4. 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 logn) time complexity?

        这个题目要求找到最长递增子序列的长度，用动态规划很好解决，建立一个数组，储存到每个节点递增子序列的长度，对于每一个节点更新其后所有比它大的节点到此节点的最大距离，当所有节点更新完，找出最大值就是最长递增子序列的长度。

class Solution {public:    int lengthOfLIS(vector<int>& nums) {        int s = nums.size();        if(!s) return 0;    vector<int> L(s);    int Max = 1;    for (int i = 0; i < s - 1; i++) {    if (L[i] == 0) L[i]=1;    for (int j = i + 1; j < s; j++) {    if (nums[i]<nums[j]&&L[j] < L[i] + 1) L[j] = L[i] + 1;    if (Max < L[j]) Max = L[j];    }    }    return Max;    }};