Leetcode_300_ rise longest sequence

Title description: given an unordered integer array, the length of the longest found rising sequence.
Example:

输入: [10,9,2,5,3,7,101,18]
输出: 4 
解释: 最长的上升子序列是 [2,3,7,101]，它的长度是 4。

ALGORITHM: Title rise longest sequence requirements, need to set up a DP [] array for recording the final result length, dp [i] is the last element contains the sequence nums [i] of length. dp [i] values are only two cases, the first case: the current surface element are not currently nums [i] value of the element is small, dp [i] = 1, i.e., contains only elements themselves, the length is 1. Second case: Present surface of the small element ratio nums [i], the current will nums [i] is added to the back of the element is smaller than nums [i], dp [i ] = dp [i] +1.
Specifically: initializing the array dp, dp [i] = 1, ans = -1 is used to return the maximum length increment, when nums [i]> when nums [j], and dp [i] <dp [j ] +1 when , updated dp [i] = dp [j ] +1, DP calculate a [i], to update the look ans, ans = max (ans, dp [i]), the maximum final ans is increasing subsequence length
i

Code:

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