LeetCode - Longest Increasing Subsequence

解法一  DP O(n^2)

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        if(nums.empty()) return 0;
        int n = nums.size();
        vector<int> dp(n, 1);
        int res = 1;
        for(int i=0;i<n;i++){
            for(int j=0;j<i;j++){
                if(nums[j]<nums[i]) dp[i] = max(dp[j]+1, dp[i]);
                res = max(res, dp[i]);
            }
        }
        return res;
    }
};

解法二  binary search O(n logn)

如何利用binary search找Longest Increasing Subsequence？
维护一个vector，每一个元素代表 increasing substring position i 上最小的数
update: 小于首元素代替，大于末元素加在末尾，大于首元素小于末元素代替第一个比它大的元素
注：这个维护的vector并非最后的increasing substring，只是元素个数相同

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        int n = nums.size();
        vector<int> dp;
        for(int i=0;i<n;i++){
            int t = nums[i];
            int l=0, r=dp.size();
            while(l<r){
                int mid = l+(r-l)/2;
                if(t>dp[mid]) l=mid+1;
                else r=mid;
            }
            if(r==dp.size()) dp.push_back(t);
            else dp[r] = t;
        }
        return dp.size();
    }
};

Reference:

http://www.cnblogs.com/grandyang/p/4938187.html

https://www.jiuzhang.com/solution/longest-increasing-subsequence/#tag-highlight-lang-java

https://blog.csdn.net/left_la/article/details/11951085