【LeetCode & 劍指offer刷題】動態規劃與貪婪法題10:Longest Increasing Subsequence
阿新 • • 發佈:2019-01-06
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//問題:找最長遞增子序列(可以不要求連續) /* 方法:動態規劃 dp[i]表示以nums[i]為結尾的最長遞增子串的長度 dp[i] = max(dp[i], dp[j] + 1)
O(n^2) O(n)
*/
class Solution
{
public:
int lengthOfLIS(vector<int>& nums)
{
vector<int> dp(nums.size(), 1); //dp值初始化為1,dp[0] = 1,一個元素長度為1
int res = 0;
for (int i = 0; i < nums.size(); i++) //遍歷數組,以num[i]結尾
{
for (int j = 0; j < i; j++) //遍歷num[i]以前的數(i=0~n-1,j=0~i-1)
{
if (nums[j] < nums[i] )//當遇到遞增對時,dp[j]+1,更新dp[i]
dp[i] = max(dp[i], dp[j] + 1);
}
res = max(res, dp[i]); //選擇dp[i]中的最大值,因為不確定以哪個num[i]結尾的遞增子序列最長
}
return res;
}
};
/*
掌握
* 方法:動態規劃+二分查找
* 具體過程:dp存最長遞增子序列
註意:數組dp不一定是真實的LIS,只是長度一致
* 例:
input: [0, 8, 4, 12, 2]
dp: [0]
dp: [0, 8]
dp: [0, 4]
dp: [0, 4, 12]
dp: [0 , 2, 12] which is not the longest increasing subsequence, but length of dp array results in length of Longest Increasing Subsequence.
* O(nlogn) O(n)
類似問題: Increasing Triplet Subsequence
*/
#include <algorithm>
class Solution
{
public:
int lengthOfLIS(vector<int>& a)
{
if(a.empty()) return 0;
vector<int> dp;
for (int ai : a)
{
//lower_bound返回第一個大於等於ai的位置,函數參數為(first,last) last指向區間末尾位置
//在dp[first,last)序列中尋找ai可以滿足增序插入的位置,如果找不到,說明ai比區間所有值大,返回last
//因為dp維護成一個增序數組,故可用二分查找法
auto it = lower_bound(dp.begin(), dp.end(), ai); //查找第一個大於等於ai的位置(查找ai可以插入的位置)
if (it == dp.end()) //如果區間中不存在,則push到末尾
dp.push_back(ai);
else //如果存在,則替換對應位置的元素
*it = ai;
}
return dp.size();
}
};
【LeetCode & 劍指offer 刷題筆記】目錄(持續更新中...)
Longest Increasing Subsequence
Given an unsorted array of integers, find the length of longest increasing subsequence. Example: 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- 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.
//問題:找最長遞增子序列(可以不要求連續) /* 方法:動態規劃 dp[i]表示以nums[i]為結尾的最長遞增子串的長度 dp[i] = max(dp[i], dp[j] + 1)
【LeetCode & 劍指offer刷題】動態規劃與貪婪法題10:Longest Increasing Subsequence