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?

1、(n * n)

int lengthOfLIS(int* nums, int numsSize) {
    int max = 0;
    //memorize a array
    if(numsSize == 0)
        return 0;
    int *result = malloc(numsSize * sizeof(int));
    result[0] = 1;
    max = 1;
    //go over the nums
    for(int i = 1; i < numsSize; i++){
        //check forwards 
        int j = i;
        int fwd_max = 0;
        for(; j >= 0; j--){
            //find one less than this one, and get the max fwd 
            if(nums[j] < nums[i] && fwd_max < result[j]){
                fwd_max = result[j];
    //if get the max, break;
    if(fwd_max == max)
        break;
            }
        }     
        result[i] = fwd_max + 1;
        if(result[i] > max)
            max = result[i];
    }
    return max;
}

2、（nlgn）

int lengthOfLIS(vector<int>& nums) {
    vector<int> res;
    for(int i=0; i<nums.size(); i++) {
        auto it = std::lower_bound(res.begin(), res.end(), nums[i]);
        if(it==res.end()) res.push_back(nums[i]);
        else *it = nums[i];
    }
    return res.size();
}

• 维持一个从小到大的数组
• 如果在数组中检测到有比它大的值就用它替换第一个比它大的值