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 is

4

. 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;
}
};