[leetcode] 300. Longest Increasing Subsequence (Medium)
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题意:
求最长增长的子序列的长度。
思路:
利用DP存取以i作为最大点的子序列长度。
Runtime: 20 ms, faster than 35.21% of C++ online submissions for Longest Increasing Subsequence.
class Solution { public: int lengthOfLIS(vector<int> &nums) { if (nums.size() == 0) return 0; vector<int> dp(nums.size(), 0); dp[0] = 1; int resNum = 1; for (int i = 1; i < nums.size(); i++) { int curmaxNum = 0; for (int j = 0; j < i; j++) { if (nums[i] > nums[j]) curmaxNum = max(dp[j], curmaxNum); } dp[i] = curmaxNum + 1; resNum = max(dp[i], resNum); } return resNum; } };
解法二:
讨论区里的最优解:
利用一个容器去动态存储一个增长子序列,遍历Nums,对每一个nums[i],在容器中寻找是否有大于等于nums[i]的元素,若存在,则将改值替换为nums[i];
若不存在,则将nums[i]加入该容器,于是容器中的元素永远是 小于 关系的。
0ms.
class Solution { public: int lengthOfLIS(vector<int>& nums) { vector<int> res; for(int i=0; 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(); } };
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