300. Longest Increasing Subsequence

Given an unsorted array of integers, find the length of longest increasing subsequence.

Example:

Input: [10,9,2,5,3,7,101,18]
Output: 4 
Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4. 

Note:

• 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?

最长递增子序列

C++(4ms):二分

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        vector<int> tails(nums.size() , 0) ;
        int len = 0 ;
        for(int num : nums){
            int index = binarySearch(tails,len,num) ;
            tails[index] = num ;
            if (index == len){
                len++ ;
            }
        }
        return len ;
    }
    
    int binarySearch(vector<int> tails , int len , int num){
        int left = 0 ;
        int right = len ;
        while(left < right){
            int mid = left + (right - left) / 2 ;
            if (tails[mid] == num){
                return mid ;
            }else if(tails[mid] < num){
                left = mid + 1 ;
            }else{
                right = mid ;
            }
        }
        return left ;
    }
};

C++（20ms）：dp

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