[LeetCode] Longest Increasing Subsequence 最长递增子序列的长度

声明：原题目转载自LeetCode，解答部分为原创

Problem ：

    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?

Solution：

    思路：假定f（k）为数组前k位中最长递增子序列的长度，状态转换方程为

        f（k） =  max { max( f（i）+ 1， f（k）) }，其中 i 为满足条件" i <= k && array[ i ] < array[ k ]" 的所有值

        时间复杂度为O（n ^ 2），代码如下：

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