给出一个无序的整形数组,找到最长上升子序列的长度。
例如,
给出 [10, 9, 2, 5, 3, 7, 101, 18],
最长的上升子序列是 [2, 3, 7, 101],因此它的长度是4。因为可能会有超过一种的最长上升子序列的组合,因此你只需要输出对应的长度即可。
你的算法的时间复杂度应该在 O(n2) 之内。
进阶: 你能将算法的时间复杂度降低到 O(n log n) 吗?

详见:https://leetcode.com/problems/longest-increasing-subsequence/description/

Java实现:

class Solution {
public int lengthOfLIS(int[] nums) {
int n=nums.length;
if(n==0||nums==null){
return 0;
}
int[] dp=new int[n];
Arrays.fill(dp,1);
int res=1;
for(int i=1;i<n;++i){
for(int j=0;j<i;++j){
if(nums[j]<nums[i]&&dp[j]+1>dp[i]){
dp[i]=dp[j]+1;
}
if(res<dp[i]){
res=dp[i];
}
}
}
return res;
}
}

C++实现:

方法一:

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

方法二:

class Solution {
public:
int lengthOfLIS(vector<int>& nums) {
int size=nums.size();
if(size==0||nums.empty())
{
return 0;
}
vector<int> res;
for(int i=0;i<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();
}
};

参考:https://www.cnblogs.com/grandyang/p/4938187.html

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