Longest Increasing Subsequence

最新推荐文章于 2024-09-11 15:54:46 发布

原创最新推荐文章于 2024-09-11 15:54:46 发布 · 194 阅读

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本文介绍了一种求解最长递增子序列(LIS)问题的算法，包括O(n^2)的时间复杂度解法及更高效的O(nlogn)解法。通过动态规划的方法，实现了对整数数组中最长递增子序列长度的计算。

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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 is 4. 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 log n) time complexity?

解：

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        const int n=nums.size();
        if(n==0)return 0;
        vector<int> nls(n);  //数组保存以当前位置为开头元素到数列末尾的最长递增序列长度
        int res=0;

        for(int i=n-1;i>=0;--i)
        {

            int cmax=0;
            for(int j=i+1;j<n;++j)
            {
                if(nums[j]>nums[i])cmax=max(cmax,nls[j]);
            }
            nls[i]=cmax+1;
            res=max(res,nls[i]);
        }

        return res;

    }
};

上面dp时间复杂度是n平方，
下面改变dp数组的含义可以得到nlogn的解法

public class Solution {
    public int lengthOfLIS(int[] nums) {            
        int[] dp = new int[nums.length];//这里数组i位置保存的是长度为i-1的LIS末尾值的最小值
        int len = 0;

        for(int x : nums) {
            int i = Arrays.binarySearch(dp, 0, len, x);
            if(i < 0) i = -(i + 1);
            dp[i] = x;
            if(i == len) len++;
        }

        return len;
    }
}