[Oracle] LeetCode 300 Longest Increasing Subsequence

Given an integer array nums, return the length of the longest strictly increasing subsequence.

A subsequence is a sequence that can be derived from an array by deleting some or no elements without changing the order of the remaining elements. For example, [3,6,2,7] is a subsequence of the array [0,3,1,6,2,2,7].

Solution

很直接的 \(DP\) 复杂度为 \(O(N^2)\), 即设 \(dp[i]\) 表示以 \(i\) 结尾序列的答案，那么转移为：

\[dp[i]=\max(dp[i], dp[j]+1), \ j< i \]

且条件是 \(nums[j]<nums[i]\)

考虑如何变成 \(O(N\log N)\) 的做法，每次当下一个数比当前列表中最后一个大时，那么直接加在后面；否则利用 \(lower\_bound\) 进行替换，因为我们只关心最大的长度

点击查看代码

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        int len=0;
        for(auto ele:nums){
            if(len==0 || nums[len-1]<ele){
                nums[len] = ele;len++;
            }
            else {
                *lower_bound(nums.begin(), nums.begin()+len, ele)=ele;
            }
        }
        return len;
    }
};