Longest Increasing Subsequence

问题：

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 is4. Note that there may be more than one LIS combination, it is only necessary for you to return the length.

思路：

    该题采用动态规划的思想来解决，用F（i）表示以数组i处元素结尾的最长递增子序列，则F（i + 1） = max(F(k) + 1), (其中k在0~i之间，且数组的第k个元素小于第i + 1 个元素)。

代码：

class Solution {public:    int lengthOfLIS(vector<int>& nums) {        int size = nums.size();        if(size < 1){            return 0;        }                int *cache = new int[size];        for(int i = 0; i < size; i++){            cache[i] = 0;        }                cache[0] = 1;        int result = 1;        for(int i = 1; i < size; i++){            int maxsize = 1;            for(int j = i - 1; j >= 0; j--){                if(nums[i] > nums[j]){                    if(cache[j] + 1 > maxsize){                        maxsize = cache[j] + 1;                    }                                     }            }            cache[i] = maxsize;            if(cache[i] > result){                result = cache[i];            }        }        return result;    }};