300. Longest Increasing Subsequence

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#算法分析与设计 #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中最长递增子序列的有效查找。

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) {
        int size = nums.size();
        if (size == 0) return 0; 
        vector<int> len(size, 1);
        int res = 1;
        for (int i = 1; i < size; ++i) {
            for (int j = 0; j < i; ++j) {
                if (nums[j] < nums[i]) len[i] = max(len[i], len[j]+1);
            }//for
            res = max (res, len[i]);
        }//for
        return res;
    }
};