Leetcode 300. Longest Increasing Subsequence

vincentCheungM

于 2017-04-08 16:50:45 发布

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本文介绍了解决LeetCode 300题——最长递增子序列的方法，通过动态规划实现，详细阐述了算法的实现过程，并提供了一个C++示例代码。该算法的时间复杂度为O(n²)，适用于寻找给定整数序列中最长的递增子序列。

Leetcode 300. Longest Increasing Subsequence

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题目描述

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.

输入：一组序列
输出：最长不减子序列的长度

思路

动态规划：用L(j)表示已第j个数结尾的最长递增子数列的长度，
可得到下面的算法。复杂度O(n^2)

for j = 1, 2, ..., n:
    L(j) = 1 + max{L(i):(i,j)∈E}
return max L(j)

代码

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        if(nums.size()==0) return 0;
        int L[nums.size()];

        int maxi;
        for(int j=0;j<nums.size();j++){
            maxi = 0;
            for(int i =0;i<=j;i++){
                //if (i,j)∈E
                if(nums[i]<nums[j]){
                    maxi = L[i]>maxi?L[i]:maxi;
                }
                L[j] = 1+maxi;
            }
        }

        int leng = 0;
        for(int j =0;j<nums.size();j++){
            leng = max(leng,L[j]);
        }
        return leng;
    }
};