题目：
Given a sequence of integers, find the longest increasing subsequence (LIS).
You code should return the length of the LIS.
Clarification
What's the definition of longest increasing subsequence?
The longest increasing subsequence problem is to find a subsequence of a given sequence in which the subsequence's elements are in sorted order, lowest to highest, and in which the subsequence is as long as possible. This subsequence is not necessarily contiguous, or unique.
https://en.wikipedia.org/wiki/Longest_increasing_subsequence
Example
For [5, 4, 1, 2, 3], the LIS is [1, 2, 3], return 3
For [4, 2, 4, 5, 3, 7], the LIS is [2, 4, 5, 7], return 4
题解：
For dp[i], dp[i] is max(dp[j]+1, dp[i]), for all j < i and nums[j] < nums[i].
Solution 1 ()
class Solution {
public:
int longestIncreasingSubsequence(vector<int> nums) {
if (nums.empty()) {
return ;
}
vector<int> dp(nums.size(), );
int res = ;
for (int i = ; i < nums.size(); ++i) {
for (int j = ; j < i; ++j) {
if (nums[j] < nums[i]) {
dp[i] = max(dp[i], dp[j] + );
}
}
res = max(dp[i], res);
}
return res;
}
};
Solution 2 ()
class Solution {
public:
/**
* @param nums: The integer array
* @return: The length of LIS (longest increasing subsequence)
*/
int longestIncreasingSubsequence(vector<int> nums) {
vector<int> res;
for(int i=; i<nums.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();
}
};
Solution 3 ()
class Solution {
public:
int longestIncreasingSubsequence(vector<int> nums) {
if (nums.empty()) {
return ;
}
vector<int> tmp;
tmp.push_back(nums[]);
for (auto num : nums) {
if (num < tmp[]) {
tmp[] = num;
} else if (num > tmp.back()) {
tmp.push_back(num);
} else {
int begin = , end = tmp.size();
while (begin < end) {
int mid = begin + (end - begin) / ;
if (tmp[mid] < num) {
begin = mid + ;
} else {
end = mid;
}
}
tmp[end] = num;
}
}
return tmp.size();
}
};