300. Longest Increasing Subsequence

300. Longest Increasing Subsequence

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?

題意：給定一個無序序列，求最長遞增子序列（不能含有相等），結果只要求返回最大長度。

分析：每一個數字都是一個只含一個數的子序列，後一個數字可以通過前一個數字的最長遞增子序列來求，所以可以用動態規劃來做。時間複雜度是O(n2)  o(╥﹏╥)o 

C++代碼：

class Solution {
public:
        int lengthOfLIS(vector<int>& nums) {
         int length = nums.size();
        if(length <= 1)
            return length;
        else
        {
            int* p = new int[length+1];
            int max = 1;
            for(int i = 0;i <length;i++)
                p[i] = 1;
            for(int i = 1;i < length;i++)
                for(int j = 0;j < i;j++)
                {
                    if(nums[i] > nums[j]&&p[i] < p[j]+1)
                    {
                        p[i] = p[j] + 1;
                        max = p[i] > max ? p[i] : max;
                    }
                }
            return max;
        }
    }
};