leetcode:Longest Increasing Subsequence

原创于 2016-07-05 10:36:31 发布 · 448 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

本文介绍了一种求解最长递增子序列长度的算法，该算法的时间复杂度为O(n²)，通过动态规划的方式实现了对未排序整数数组的处理。举例说明了如何寻找给定数组中的最长递增子序列及其长度。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

class Solution {
public:
	int lengthOfLIS(vector<int>& nums) {
	    if(nums.empty())
	    	return 0;
		int size = nums.size();
		vector<int> length(size,1);

		for(int i=1;i<size;i++)
		{
			int largest = 0;
			int len = 0;
			for(int j=0;j<i;j++)
			{
				if(nums[i]>nums[j])
				{
					len = length[j]+1;
				}
				else{
					len = 1;
				}

				if(len>largest)
				{
					largest = len;
				}
			}
			length[i] = largest;
		}

		int l =0;
		vector<int>::iterator it = length.begin();

		for(;it!=length.end();it++)
		{
			if(*it>l)
				l=*it;
		}

		return l;
	}
};