300. Longest Increasing Subsequence

最新推荐文章于 2025-08-08 11:52:34 发布

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最新推荐文章于 2025-08-08 11:52:34 发布

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分类专栏： # 算法实践文章标签： DP c++

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本文介绍了一种求解最长递增子序列(LIS)问题的方法，通过动态规划算法实现，复杂度为O(n^2)，并提出了使用二分搜索优化至O(n log n)的解决方案。

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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?

题意：求数组的最长递增子序列

思路：dp[i] 记录以nums[i]结尾的递增序列的长度，初始化dp[0] = 1; 则dp[i+1]则是比较nums[i+1] 与之前所有的nums[j]的值，并根据dp[j]更新dp[i+1],没有比nums[i+1]小的则dp[i+1] = 1,所以可以先初始化所有的dp[i]=1。复杂度为O(n^2)。动态规划+binary search可是复杂度降为O(nlogn)。

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