leetcode链接:https://leetcode-cn.com/problems/longest-increasing-subsequence/submissions/

题目描述:给定一个长度为N的序列nums[9] = {5, 6, 7, 1, 2, 8}

解法一:动态规划(O(n^2))

f[i]表示以第i个数字结尾的最长上升序列的长度。

如果nums[i]比前面某个值大,那么就比较一下f[i]和f[j]+1,取最大即可。

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

解法二:利用二分优化成O(n*logn)

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        int n = nums.size();
        if(n == 0) return 0;
        vector<int> B(n+1);
        B[0] = nums[0];
        int len = 1, pos = 0;
        for(int i = 1; i < n; i++){
            if(nums[i] > B[len-1]){
                B[len] = nums[i];
                len++;
            }
            else{
                pos = BiSearch(B, len, nums[i]);
                B[pos] = nums[i];
            }
        }
        return len;
    }

    int BiSearch(vector<int>& B, int len, int w){
        int left = 0, right = len-1;
        int mid;
        while(left <= right){
            mid = left + (right - left) / 2;
            if(B[mid] > w)
                right = mid - 1;
            else if(B[mid] < w)
                left = mid + 1;
            else
                return mid;
        }
        return left;
    }
};

 

08-13 13:26