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

Given an unsorted array of integers, find the length of longest increasing subsequence.

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.

Your algorithm should run in O(n2) complexity.

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

 

解法1: 动态规划DP,类似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)

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,需要遍历在j之前的所有位置i(从0到j-1),找出满足条件a[i]<a[j]的L(i),求出max(L(i))+1即为L(j)的值。最后,我们遍历所有的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)。

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。所以该数组最长递增子序列长度为4,序列为{5,6,7,8}。

解法2: 二分查找法Binary Search, 维护一个单调递增子序列,如果当前值小于单调递增子序列中的某个元素,则替换之,因为单调递增子序列能否增长,值取决于最后一个元素,替换内部的元素并不影响。Time: O(nlogn)

假设存在一个序列d[1..9] = { 2,1 ,5 ,3 ,6,4, 8 ,9, 7},可以看出来它的LIS长度为5。
定义一个序列B,然后让 i = 1 to 9 逐个考察这个序列。用一个变量Len来记录现在最长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
然后,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],所以令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,因为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,于是可以把5淘汰掉,这时候B[1..2] = 1, 3,Len = 2
继续,d[5] = 6,它在3后面,因为B[2] = 3, 而6在3后面,于是很容易可以推知B[3] = 6, 这时B[1..3] = 1, 3, 6,还是很容易理解吧? Len = 3 了噢。
第6个, d[6] = 4,你看它在3和6之间,于是我们就可以把6替换掉,得到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。于是B[4] = 8。Len变成4了
第8个, d[8] = 9,得到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,所以我们知道,最新的B[4] =7,B[1..5] = 1, 3, 4, 7, 9,Len = 5。
于是我们知道了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,我们就可以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,但是如果后面再出现两个数字 8 和 9,那么就可以把8更新到d[5], 9更新到d[6],得出LIS的长度为6。
然后发现一件事情:在B中插入数据是有序的,而且是进行替换而不需要挪动——也就是说,可以使用二分查找,将每一个数字的插入时间优化到O(logN)~~~~~于是算法的时间复杂度就降低到了O(NlogN)~!

Java: DP

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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posted @ 2018-03-09 10:15  轻风舞动  阅读(1007)  评论(0编辑  收藏  举报