题目
Given an unsorted array of integers, find the length of longest increasing subsequence.
Example 1:
Input: [10,9,2,5,3,7,101,18]
Output: 4
Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.
Note:
- 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.
算法思路
本周继续学习动态规划的问题,本题要求实现书本上动态规划里面的经典问题——求最长增长子序列。
我们拆解成子问题,从左到右依次计算以每个元素结尾的最长增长子序列的长度,而计算当前元素的最长增长子序列的长度时,只需要在前面比当前元素小的元素中选取计算好的最长的长度加一即可。
分解子问题的状态方程
L(i) = 1+max{L(j) : j<i且S[j] < S[i]}
L[i]指的是以第i个元素结束的最长增长子序列的长度,S[i]指序列中的第i个元素
C++代码
class Solution {
public:
int lengthOfLIS(vector<int>& nums) {
if (nums.size() == 0)
{
return 0;
}
int longest[nums.size()];
longest[0] = 1;
for (int i = 1; i < nums.size(); ++i)
{
int max = 0;
for (int j = 0; j < i; ++j)
{
if (nums[i] > nums[j] && max < longest[j])
{
max = longest[j];
}
}
longest[i] = max + 1;
}
int result = 0;
for (int i = 0; i < nums.size(); ++i)
{
if (result < longest[i])
{
result = longest[i];
}
}
return result;
}
};