[LeetCode] 300. Longest Increasing Subsequence 最长递增子序列

2021年09月15日 阅读数:2
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Given an unsorted array of integers, find the length of longest increasing subsequence.html

For example,
Given [10, 9, 2, 5, 3, 7, 101, 18],
The longest increasing subsequence is [2, 3, 7, 101], therefore the length is 4. Note that there may be more than one LIS combination, it is only necessary for you to return the length.java

Your algorithm should run in O(n2) complexity.python

Follow up: Could you improve it to O(n log n) time complexity?算法

 

解法1: 动态规划DP,相似brute force的解法,维护一个一维dp数组,其中dp[i]表示以nums[i]为结尾的最长递增子串的长度。Time: O(n^2)数组

设长度为N的数组为{a0,a1, a2, ...an-1),则假定以aj结尾的数组序列的最长递增子序列长度为L(j),则L(j)={ max(L(i))+1, i<j且a[i]<a[j] }。也就是说,须要遍历在j以前的全部位置i(从0到j-1),找出知足条件a[i]<a[j]的L(i),求出max(L(i))+1即为L(j)的值。最后,咱们遍历全部的L(j)(从0到N-1),找出最大值即为最大递增子序列。时间复杂度为O(N^2)。app

例如给定的数组为{5,6,7,1,2,8},则L(0)=1, L(1)=2, L(2)=3, L(3)=1, L(4)=2, L(5)=4。因此该数组最长递增子序列长度为4,序列为{5,6,7,8}。ide

解法2: 二分查找法Binary Search, 维护一个单调递增子序列,若是当前值小于单调递增子序列中的某个元素,则替换之,由于单调递增子序列可否增加,值取决于最后一个元素,替换内部的元素并不影响。Time: O(nlogn)post

假设存在一个序列d[1..9] = { 2,1 ,5 ,3 ,6,4, 8 ,9, 7},能够看出来它的LIS长度为5。
定义一个序列B,而后让 i = 1 to 9 逐个考察这个序列。用一个变量Len来记录如今最长LIS长度。
首先,把d[1]有序地放到B里,令B[1] = 2,就是说当只有1一个数字2的时候,长度为1的LIS的最小末尾是2。这时Len=1
而后,把d[2]有序地放到B里,令B[1] = 1,就是说长度为1的LIS的最小末尾是1,d[1]=2已经没用了,很容易理解吧。这时Len=1
接着,d[3] = 5,d[3]>B[1],因此令B[1+1]=B[2]=d[3]=5,就是说长度为2的LIS的最小末尾是5,很容易理解吧。这时候B[1..2] = 1, 5,Len=2
再来,d[4] = 3,它正好加在1,5之间,放在1的位置显然不合适,由于1小于3,长度为1的LIS最小末尾应该是1,这样很容易推知,长度为2的LIS最小末尾是3,因而能够把5淘汰掉,这时候B[1..2] = 1, 3,Len = 2
继续,d[5] = 6,它在3后面,由于B[2] = 3, 而6在3后面,因而很容易能够推知B[3] = 6, 这时B[1..3] = 1, 3, 6,仍是很容易理解吧? Len = 3 了噢。
第6个, d[6] = 4,你看它在3和6之间,因而咱们就能够把6替换掉,获得B[3] = 4。B[1..3] = 1, 3, 4, Len继续等于3
第7个, d[7] = 8,它很大,比4大,嗯。因而B[4] = 8。Len变成4了
第8个, d[8] = 9,获得B[5] = 9,嗯。Len继续增大,到5了。
最后一个, d[9] = 7,它在B[3] = 4和B[4] = 8之间,因此咱们知道,最新的B[4] =7,B[1..5] = 1, 3, 4, 7, 9,Len = 5。
因而咱们知道了LIS的长度为5。
注意,这个1,3,4,7,9不是LIS,它只是存储的对应长度LIS的最小末尾。有了这个末尾,咱们就能够一个一个地插入数据。虽然最后一个d[9] = 7更新进去对于这组数据没有什么意义,可是若是后面再出现两个数字 8 和 9,那么就能够把8更新到d[5], 9更新到d[6],得出LIS的长度为6。
而后发现一件事情:在B中插入数据是有序的,并且是进行替换而不须要挪动——也就是说,可使用二分查找,将每个数字的插入时间优化到O(logN)~~~~~因而算法的时间复杂度就下降到了O(NlogN)~!优化

Java: DPurl

public class Solution {  
    public int lengthOfLIS(int[] nums) {  
        if (nums == null || nums.length == 0) return 0;  
        int max = 1;  
        int[] lens = new int[nums.length];  
        Arrays.fill(lens, 1);  
        for(int i=1; i<nums.length; i++) {  
            for(int j=0; j<i; j++) {  
                if (nums[j]<nums[i]) lens[i] = Math.max(lens[i], lens[j]+1);  
            }  
            max = Math.max(max, lens[i]);  
        }  
        return max;  
    }  
}  

Java:  BS

public class Solution {  
    public int lengthOfLIS(int[] nums) {  
        int[] increasing = new int[nums.length];  
        int size = 0;  
        for(int i=0; i<nums.length; i++) {  
            int left=0, right=size-1;  
            while (left<=right) {  
                int m=(left+right)/2;  
                if (nums[i] > increasing[m]) left = m + 1;  
                else right = m - 1;  
            }  
            increasing[left] = nums[i];  
            if (left==size) size ++;  
        }  
        return size;  
    }  
}  

Java: TreeSet

public class Solution {  
    public int lengthOfLIS(int[] nums) {  
        if (nums == null || nums.length == 0) return 0;  
        int max = 1;  
        TreeSet<Integer> ts = new TreeSet<>(new Comparator<Integer>() {  
            @Override  
            public int compare(Integer i1, Integer i2) {  
                return Integer.compare(nums[i1], nums[i2]);  
            }  
        });  
        int[] lens = new int[nums.length];  
        Arrays.fill(lens, 1);  
        for(int i=0; i<nums.length; i++) {  
            if (ts.contains(i)) ts.remove(i);  
            ts.add(i);  
            Set<Integer> heads = ts.headSet(i);  
            for(int head: heads) {  
                lens[i] = Math.max(lens[i], lens[head] + 1);  
            }  
            max = Math.max(max, lens[i]);  
        }  
        return max;  
    }  
}  

Python: BS, T: O(nlogn), S: O(n)

class Solution(object):
    def lengthOfLIS(self, nums):
        LIS = []
        def insert(target):
            left, right = 0, len(LIS) - 1
            # Find the first index "left" which satisfies LIS[left] >= target
            while left <= right:
                mid = left + (right - left) / 2
                if LIS[mid] >= target:
                    right = mid - 1
                else:
                    left = mid + 1
            # If not found, append the target.
            if left == len(LIS):
                LIS.append(target);
            else:
                LIS[left] = target

        for num in nums:
            insert(num)

        return len(LIS)

Python: DP, T: O(n^2), S: O(n)

class Solution(object):
    def lengthOfLIS(self, nums):
        dp = []  # dp[i]: the length of LIS ends with nums[i]
        for i in xrange(len(nums)):
            dp.append(1)
            for j in xrange(i):
                if nums[j] < nums[i]:
                    dp[i] = max(dp[i], dp[j] + 1)

        return max(dp) if dp else 0

Python: wo

class Solution(object):
    def lengthOfLIS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums:
            return 0
        n = len(nums)
        dp = [1] * n
        max_len = 1
        for i in xrange(n):
            for j in xrange(i):
                if nums[i] > nums[j]:
                    dp[i] = max(dp[i], dp[j] + 1)
                    max_len = max(max_len, dp[i])
                              
        return max_len   

C++:DP

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

C++:BS

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        if (nums.empty()) return 0;
        vector<int> ends{nums[0]};
        for (auto a : nums) {
            if (a < ends[0]) ends[0] = a;
            else if (a > ends.back()) ends.push_back(a);
            else {
                int left = 0, right = ends.size();
                while (left < right) {
                    int mid = left + (right - left) / 2;
                    if (ends[mid] < a) left = mid + 1;
                    else right = mid;
                }
                ends[right] = a;
            }
        }
        return ends.size();
    }
}; 

C++:BS

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        vector<int> dp;
        for (int i = 0; i < nums.size(); ++i){

            int lo = 0, hi = dp.size();
            while (lo < hi){
                int mi = (lo + hi)/2;
                if (dp[mi] < nums[i]) lo = mi + 1;
                else hi = mi;
            }
            if (hi == dp.size())
                dp.push_back(nums[i]);
            else dp[hi] = nums[i];
        }
        return dp.size();
    }
};

   

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